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Which method to use to integrate this function?
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up vote
14
down vote
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Recently I've been working on problems including integration and I'm having some difficulties solving this integral :
$$int_0^1 (1-x)sqrt4x-x^2 dx$$
I tried to rewrite it in this form:
$$int_0^1 (1-x)xsqrtfrac4-xx dx$$ and then subsitute $frac4-xx=t^2$ and by this substitution I get a function way easier integrated of this form:
$$-32intfrac(t^2-3)t^2(1+t^2)^4dt$$
I can integrate this function easily by simplifying it but the problem is that I cannot define its borders because $t$ is not defined for $x=0$ so this means that my substitution doesn't hold. But I don't have any other idea how to solve it so any help will be appreciated.Thank you!
real-analysis integration definite-integrals
asked yesterday
Maths Survivor
800119
add a comment |Â
up vote
14
down vote
favorite
Recently I've been working on problems including integration and I'm having some difficulties solving this integral :
$$int_0^1 (1-x)sqrt4x-x^2 dx$$
I tried to rewrite it in this form:
$$int_0^1 (1-x)xsqrtfrac4-xx dx$$ and then subsitute $frac4-xx=t^2$ and by this substitution I get a function way easier integrated of this form:
$$-32intfrac(t^2-3)t^2(1+t^2)^4dt$$
I can integrate this function easily by simplifying it but the problem is that I cannot define its borders because $t$ is not defined for $x=0$ so this means that my substitution doesn't hold. But I don't have any other idea how to solve it so any help will be appreciated.Thank you!
real-analysis integration definite-integrals
asked yesterday
Maths Survivor
800119
5
I'd write $4x-x^2=4-(x-2)^2$ and write $x-2=2sintheta$.
– Lord Shark the Unknown
yesterday
the domain of integration $xin(0,1)$ should be transformed into $tin(3,infty)$ with a reversal of orientation.
– robjohn♦
yesterday
add a comment |Â
up vote
14
down vote
favorite
up vote
14
down vote
favorite
Recently I've been working on problems including integration and I'm having some difficulties solving this integral :
$$int_0^1 (1-x)sqrt4x-x^2 dx$$
I tried to rewrite it in this form:
$$int_0^1 (1-x)xsqrtfrac4-xx dx$$ and then subsitute $frac4-xx=t^2$ and by this substitution I get a function way easier integrated of this form:
$$-32intfrac(t^2-3)t^2(1+t^2)^4dt$$
I can integrate this function easily by simplifying it but the problem is that I cannot define its borders because $t$ is not defined for $x=0$ so this means that my substitution doesn't hold. But I don't have any other idea how to solve it so any help will be appreciated.Thank you!
real-analysis integration definite-integrals
asked yesterday
Maths Survivor
800119
Recently I've been working on problems including integration and I'm having some difficulties solving this integral :
$$int_0^1 (1-x)sqrt4x-x^2 dx$$
I tried to rewrite it in this form:
$$int_0^1 (1-x)xsqrtfrac4-xx dx$$ and then subsitute $frac4-xx=t^2$ and by this substitution I get a function way easier integrated of this form:
$$-32intfrac(t^2-3)t^2(1+t^2)^4dt$$
I can integrate this function easily by simplifying it but the problem is that I cannot define its borders because $t$ is not defined for $x=0$ so this means that my substitution doesn't hold. But I don't have any other idea how to solve it so any help will be appreciated.Thank you!
real-analysis integration definite-integrals
real-analysis integration definite-integrals
asked yesterday
Maths Survivor
800119
asked yesterday
Maths Survivor
800119
asked yesterday
Maths Survivor
800119
asked yesterday
Maths Survivor
800119
asked yesterday
Maths Survivor
800119
800119
5
I'd write $4x-x^2=4-(x-2)^2$ and write $x-2=2sintheta$.
– Lord Shark the Unknown
yesterday
the domain of integration $xin(0,1)$ should be transformed into $tin(3,infty)$ with a reversal of orientation.
– robjohn♦
yesterday
add a comment |Â
5
I'd write $4x-x^2=4-(x-2)^2$ and write $x-2=2sintheta$.
– Lord Shark the Unknown
yesterday
the domain of integration $xin(0,1)$ should be transformed into $tin(3,infty)$ with a reversal of orientation.
– robjohn♦
yesterday
5
5
I'd write $4x-x^2=4-(x-2)^2$ and write $x-2=2sintheta$.
– Lord Shark the Unknown
yesterday
I'd write $4x-x^2=4-(x-2)^2$ and write $x-2=2sintheta$.
– Lord Shark the Unknown
yesterday
the domain of integration $xin(0,1)$ should be transformed into $tin(3,infty)$ with a reversal of orientation.
– robjohn♦
yesterday
the domain of integration $xin(0,1)$ should be transformed into $tin(3,infty)$ with a reversal of orientation.
– robjohn♦
yesterday
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
7
down vote
accepted
It is always didactically good to give a solution for the given start. I will try it, although trigonometric substitutions from the beginning may save typing.
The substitution $t=sqrt(4-x)/x$ is ok, but we have to use improper integrals. The solution would be as follows. For $x=0+$ the corresponding (limiting) value of $t$ is $t=sqrt4/0_+=+infty$, and for $x=1$ we get $t=sqrt(4-1)/1=colorredsqrt 3$. (It is square three, maybe the one reason for writing this answer.)
$$
beginaligned
J &=
int_0^1 (1-x)sqrt4x-x^2 ; dx
\
&=
int_infty^sqrt 3
-32frac(t^2-3)t^2(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
frac(t^2+2t^2+1)-5(t^2+1)+4(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
left[
frac1(1+t^2)^2
-5frac1(1+t^2)^3
+4frac1(1+t^2)^4
right];dt
\
&=32(K_2-5K_3+4K_4) ,
endaligned
$$
where $K_n$ is the integral on the same interval of $(1+t^2)^-n$. We have the recursion
$$
beginaligned
K_n
&=int_sqrt 3^inftyt'frac1(1+t^2)^n;dt
\
&=frac t(1+t^2)^nBigg|_sqrt 3^infty
-
int_sqrt 3^infty
tcdot frac-2nt(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2nint_sqrt 3^inftyfrac(t^2+1)-1(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2n(K_n-K_n+1) ,text i.e.
\
K_n+1
&=
frac 12nleft[ (2n-1)K_n-fracsqrt34^n right] .
\[2mm]
&qquadtextThis gives:
\
K_1 &=arctan infty-arctansqrt 3 =left(frac 12-frac 13right)pi
=frac 16pi ,
\
K_2 &=frac 12left[K_1-fracsqrt 34right]=frac 112pi - fracsqrt 38 ,
\
K_3 &=frac 14left[3K_2-fracsqrt 316right]
=frac 116pi - frac7sqrt 364 ,
\
K_4 &=frac 16left[5K_3-fracsqrt 364right]
=frac 596pi - frac3sqrt 332 ,
%\
%K_5 &=frac 18left[7K_4-fracsqrt 3216right]
%=frac 35768pi - frac169sqrt 32048 ,
\
&qquadtext so putting all together
\
J&=32(K_2-5K_3+4K_4)
\
&=-frac23pi+frac 3sqrt 32 .
endaligned
$$
As said, not the quick way, but all details are displayed. Computer check:
sage: var('x,t');
sage: integral( (1-x)*sqrt(4*x-x^2), x, 0, 1 ).simplify()
-2/3*pi + 3/2*sqrt(3)
sage: integral( 32*(t^2-3)*t^2/(1+t^2)^4, t, sqrt(3), oo ).simplify()
-2/3*pi + 3/2*sqrt(3)
answered yesterday
dan_fulea
4,7631211
Are we allowed to use the infinity sign like this $arctan infty$? Because a professor of mine says that infinity sign should be used only when we have limits otherwise it's wrong to use it in other(direct) ways.
– Maths Survivor
yesterday
In this case we have the limit, explicitly $$K_1=int_sqrt 3^inftyfrac dt1+t^2=lim_Mtoinftyint_sqrt 3^Mfrac dt1+t^2=lim_MtoinftyarctanBigg|_sqrt 3^M=lim_Mtoinftyarctan M -arctansqrt 3=frac 12pi-frac 13pi .$$ To avoid this complicated notation, it is simpler to write $arctan infty$ instead of the longer $lim_Mtoinftyarctan M$. (Same in the integral.) We apply a continuous function, $arctan$, on the "symbol" $infty$, this function has a limit. The substitution was dictating the use of an improper integral, one with an $infty$ as one limit.
– dan_fulea
yesterday
I get it , thank you for your answer .
– Maths Survivor
yesterday
1
@MathsSurvivor: Your professor is correct if you are working in real analysis with the usual real line, since there is no real number $∞$. In this case, you must define infinite limit and limit at infinity separately from the usual case, and then the "$infty$" in both of them are merely symbolic. However, we can 'add' $∞$ to the real line, to get the affinely extended real line, which is literally the closed interval $[-∞,∞]$, and the real line is literally the open interval $(-∞,∞)$.
– user21820
yesterday
1
$[-∞,∞]$ is not a field, but it is a compact topological space, and $sup(S),inf(S)$ exist for any subset $S$, and hence $limsup,liminf$ of any sequence always exists. If they are equal then the limit exists. These extend the usual ones, and may help to explain why the logical structure of the infinite limits and limits and infinity are so similar to the non-infinite case. You can check that $arctan$ on $(-∞,∞)$ can be continuously extended to $-∞$ and $∞$, and then it is literally correct to write $arctan(∞) = À/2$.
– user21820
yesterday
 |Â
show 6 more comments
up vote
21
down vote
The us try to get rid of the square root step-by-step:
$$begineqnarray* int_0^1(1-x)sqrtxsqrt4-x,dx&stackrelxmapsto z^2=& 2int_0^1z^2(1-z^2)sqrt4-z^2,dz\&stackrelzmapsto 2u=&32int_0^1/2u^2(1-4u^2)sqrt1-u^2,du\&stackrelumapstosintheta=&32int_0^pi/6sin^2thetacos^2theta(1-4sin^2theta),dtheta\&=&4int_0^pi/6left[-1+cos(2theta)+cos(4theta)-cos(6theta)right],dtheta.endeqnarray*$$
I guess you may take it from here.
answered yesterday
Jack D'Aurizio♦
274k32268638
add a comment |Â
up vote
6
down vote
You can use $int_a^b f(x) dx=int_a^b f(a+b-x) dx$ to get that $$I=int_0^1 x sqrt4-(1+x)^2 dx$$ You get$$I= int_0^1 (x+1)sqrt4 - (x+1)^2 dx-int_0^1 sqrt4 - (x+1)^2 dx$$ For the first one just substitute $4-(x+1)^2 =t$ to get $$frac12 int_0^3 sqrt t dt$$and the second one is a standard square root integral.
answered yesterday
Zacky
2,1111327
How did you get $sqrt4-(1+x^2)$ shouldn't it be $sqrt4(1-x)+(1-x)^2$?
– Maths Survivor
yesterday
I used that $4x-x^2 = 4-(x-2)^2 $ to simplify faster.
– Zacky
yesterday
Also $$4(1-x)-(1-x)^2 =(1-x) (4-1+x)=(1-x) (3+x)$$$$2 ^2 - (x+1)^2 =(2-1-x) (2+1+x)=(1-x)(3+x)$$
– Zacky
yesterday
Yeah it's the same , you're right
– Maths Survivor
yesterday
1
Yes, just substitute $a+b-x=trightarrow x=a+b-t$ to prove it. The new bound will be $int_b^a$, but as we have $-dx=dt$ we can get rid of that minus sign and swap those to obtain $int_a^b f(x) dx=int_a^b f(a+b-t) dt$
– Zacky
yesterday
 |Â
show 1 more comment
up vote
3
down vote
Hint:
By completing the square,
$$4x-x^2=4-(x-2)^2$$ and that hints to use the substitution
$$x-2=2cos t.$$
Then
$$int (1-x)sqrt4x-x^2dx=4int(1+2cos t)sin^2t,dt.$$
The term $sin^2t$ is integrated in the form $dfrac1-cos 2t2$, and the second term is immediate.
$$I=2t-2sin tcos t+dfrac83sin^3t=\arccosdfracx-22-(x-2)sqrt1-dfrac(x-2)^24+dfrac83left(1-dfrac(x-2)^24right)^3/2$$
answered yesterday
Yves Daoust
114k665209
add a comment |Â
up vote
3
down vote
Your approach can be completed by noting that the domain of integrtion $xin(0,1)$ gets transformed to the unbounded domain of integration $tin(sqrt3,infty)$ with reversal of orientation (which negates the integral).
We can also set $u=1-x/2implies4!left(1-u^2right)=4x-x^2$. Therefore,
$$
int_0^1(1-x)sqrt4x-x^2,mathrmdx=4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
$$
Then, setting $u=sin(theta)$ gives
$$
beginalign
4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
&=4int_pi/6^pi/2(2sin(theta)-1)overbracecos^2(theta)^1-sin^2(theta),mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00sin^3(theta)+color#090sin^2(theta)+2sin(theta)-1right)mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00frac3sin(theta)-sin(3theta)4+color#090frac1-cos(2theta)2+2sin(theta)-1right)mathrmdtheta\
%&=-left[frac23cos(3theta)+sin(2theta)+2cos(theta)+2thetaright]_pi/6^pi/2\[3pt]
%&=frac32sqrt3-frac23pi
endalign
$$
answered yesterday
robjohn♦
259k26298613
At the stage $(2s-1)c^2$, it is more direct to integrate $sc^2$ and $c^2$.
– Yves Daoust
20 hours ago
The mechanics should include $sc^n$ and $cs^n$, as well as $s^2m+1c^n$ and $c^2m+1s^n$ where you expand the even powers. Polynomials are easy.
– Yves Daoust
18 hours ago
You needn't tell me.
– Yves Daoust
18 hours ago
add a comment |Â
up vote
2
down vote
Hint: $(1-x)sqrt4x-x^2 dx=-sqrt4x-x^2dx +(2-x)sqrt4x-x^2dx=-sqrt4x-x^2dx +0.5sqrt4x-x^2d(4x-x^2)$
answered yesterday
Vasya
2,6531514
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
It is always didactically good to give a solution for the given start. I will try it, although trigonometric substitutions from the beginning may save typing.
The substitution $t=sqrt(4-x)/x$ is ok, but we have to use improper integrals. The solution would be as follows. For $x=0+$ the corresponding (limiting) value of $t$ is $t=sqrt4/0_+=+infty$, and for $x=1$ we get $t=sqrt(4-1)/1=colorredsqrt 3$. (It is square three, maybe the one reason for writing this answer.)
$$
beginaligned
J &=
int_0^1 (1-x)sqrt4x-x^2 ; dx
\
&=
int_infty^sqrt 3
-32frac(t^2-3)t^2(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
frac(t^2+2t^2+1)-5(t^2+1)+4(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
left[
frac1(1+t^2)^2
-5frac1(1+t^2)^3
+4frac1(1+t^2)^4
right];dt
\
&=32(K_2-5K_3+4K_4) ,
endaligned
$$
where $K_n$ is the integral on the same interval of $(1+t^2)^-n$. We have the recursion
$$
beginaligned
K_n
&=int_sqrt 3^inftyt'frac1(1+t^2)^n;dt
\
&=frac t(1+t^2)^nBigg|_sqrt 3^infty
-
int_sqrt 3^infty
tcdot frac-2nt(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2nint_sqrt 3^inftyfrac(t^2+1)-1(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2n(K_n-K_n+1) ,text i.e.
\
K_n+1
&=
frac 12nleft[ (2n-1)K_n-fracsqrt34^n right] .
\[2mm]
&qquadtextThis gives:
\
K_1 &=arctan infty-arctansqrt 3 =left(frac 12-frac 13right)pi
=frac 16pi ,
\
K_2 &=frac 12left[K_1-fracsqrt 34right]=frac 112pi - fracsqrt 38 ,
\
K_3 &=frac 14left[3K_2-fracsqrt 316right]
=frac 116pi - frac7sqrt 364 ,
\
K_4 &=frac 16left[5K_3-fracsqrt 364right]
=frac 596pi - frac3sqrt 332 ,
%\
%K_5 &=frac 18left[7K_4-fracsqrt 3216right]
%=frac 35768pi - frac169sqrt 32048 ,
\
&qquadtext so putting all together
\
J&=32(K_2-5K_3+4K_4)
\
&=-frac23pi+frac 3sqrt 32 .
endaligned
$$
As said, not the quick way, but all details are displayed. Computer check:
sage: var('x,t');
sage: integral( (1-x)*sqrt(4*x-x^2), x, 0, 1 ).simplify()
-2/3*pi + 3/2*sqrt(3)
sage: integral( 32*(t^2-3)*t^2/(1+t^2)^4, t, sqrt(3), oo ).simplify()
-2/3*pi + 3/2*sqrt(3)
answered yesterday
dan_fulea
4,7631211
Are we allowed to use the infinity sign like this $arctan infty$? Because a professor of mine says that infinity sign should be used only when we have limits otherwise it's wrong to use it in other(direct) ways.
– Maths Survivor
yesterday
In this case we have the limit, explicitly $$K_1=int_sqrt 3^inftyfrac dt1+t^2=lim_Mtoinftyint_sqrt 3^Mfrac dt1+t^2=lim_MtoinftyarctanBigg|_sqrt 3^M=lim_Mtoinftyarctan M -arctansqrt 3=frac 12pi-frac 13pi .$$ To avoid this complicated notation, it is simpler to write $arctan infty$ instead of the longer $lim_Mtoinftyarctan M$. (Same in the integral.) We apply a continuous function, $arctan$, on the "symbol" $infty$, this function has a limit. The substitution was dictating the use of an improper integral, one with an $infty$ as one limit.
– dan_fulea
yesterday
I get it , thank you for your answer .
– Maths Survivor
yesterday
1
@MathsSurvivor: Your professor is correct if you are working in real analysis with the usual real line, since there is no real number $∞$. In this case, you must define infinite limit and limit at infinity separately from the usual case, and then the "$infty$" in both of them are merely symbolic. However, we can 'add' $∞$ to the real line, to get the affinely extended real line, which is literally the closed interval $[-∞,∞]$, and the real line is literally the open interval $(-∞,∞)$.
– user21820
yesterday
1
$[-∞,∞]$ is not a field, but it is a compact topological space, and $sup(S),inf(S)$ exist for any subset $S$, and hence $limsup,liminf$ of any sequence always exists. If they are equal then the limit exists. These extend the usual ones, and may help to explain why the logical structure of the infinite limits and limits and infinity are so similar to the non-infinite case. You can check that $arctan$ on $(-∞,∞)$ can be continuously extended to $-∞$ and $∞$, and then it is literally correct to write $arctan(∞) = À/2$.
– user21820
yesterday
 |Â
show 6 more comments
up vote
7
down vote
accepted
It is always didactically good to give a solution for the given start. I will try it, although trigonometric substitutions from the beginning may save typing.
The substitution $t=sqrt(4-x)/x$ is ok, but we have to use improper integrals. The solution would be as follows. For $x=0+$ the corresponding (limiting) value of $t$ is $t=sqrt4/0_+=+infty$, and for $x=1$ we get $t=sqrt(4-1)/1=colorredsqrt 3$. (It is square three, maybe the one reason for writing this answer.)
$$
beginaligned
J &=
int_0^1 (1-x)sqrt4x-x^2 ; dx
\
&=
int_infty^sqrt 3
-32frac(t^2-3)t^2(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
frac(t^2+2t^2+1)-5(t^2+1)+4(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
left[
frac1(1+t^2)^2
-5frac1(1+t^2)^3
+4frac1(1+t^2)^4
right];dt
\
&=32(K_2-5K_3+4K_4) ,
endaligned
$$
where $K_n$ is the integral on the same interval of $(1+t^2)^-n$. We have the recursion
$$
beginaligned
K_n
&=int_sqrt 3^inftyt'frac1(1+t^2)^n;dt
\
&=frac t(1+t^2)^nBigg|_sqrt 3^infty
-
int_sqrt 3^infty
tcdot frac-2nt(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2nint_sqrt 3^inftyfrac(t^2+1)-1(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2n(K_n-K_n+1) ,text i.e.
\
K_n+1
&=
frac 12nleft[ (2n-1)K_n-fracsqrt34^n right] .
\[2mm]
&qquadtextThis gives:
\
K_1 &=arctan infty-arctansqrt 3 =left(frac 12-frac 13right)pi
=frac 16pi ,
\
K_2 &=frac 12left[K_1-fracsqrt 34right]=frac 112pi - fracsqrt 38 ,
\
K_3 &=frac 14left[3K_2-fracsqrt 316right]
=frac 116pi - frac7sqrt 364 ,
\
K_4 &=frac 16left[5K_3-fracsqrt 364right]
=frac 596pi - frac3sqrt 332 ,
%\
%K_5 &=frac 18left[7K_4-fracsqrt 3216right]
%=frac 35768pi - frac169sqrt 32048 ,
\
&qquadtext so putting all together
\
J&=32(K_2-5K_3+4K_4)
\
&=-frac23pi+frac 3sqrt 32 .
endaligned
$$
As said, not the quick way, but all details are displayed. Computer check:
sage: var('x,t');
sage: integral( (1-x)*sqrt(4*x-x^2), x, 0, 1 ).simplify()
-2/3*pi + 3/2*sqrt(3)
sage: integral( 32*(t^2-3)*t^2/(1+t^2)^4, t, sqrt(3), oo ).simplify()
-2/3*pi + 3/2*sqrt(3)
answered yesterday
dan_fulea
4,7631211
Are we allowed to use the infinity sign like this $arctan infty$? Because a professor of mine says that infinity sign should be used only when we have limits otherwise it's wrong to use it in other(direct) ways.
– Maths Survivor
yesterday
In this case we have the limit, explicitly $$K_1=int_sqrt 3^inftyfrac dt1+t^2=lim_Mtoinftyint_sqrt 3^Mfrac dt1+t^2=lim_MtoinftyarctanBigg|_sqrt 3^M=lim_Mtoinftyarctan M -arctansqrt 3=frac 12pi-frac 13pi .$$ To avoid this complicated notation, it is simpler to write $arctan infty$ instead of the longer $lim_Mtoinftyarctan M$. (Same in the integral.) We apply a continuous function, $arctan$, on the "symbol" $infty$, this function has a limit. The substitution was dictating the use of an improper integral, one with an $infty$ as one limit.
– dan_fulea
yesterday
I get it , thank you for your answer .
– Maths Survivor
yesterday
1
@MathsSurvivor: Your professor is correct if you are working in real analysis with the usual real line, since there is no real number $∞$. In this case, you must define infinite limit and limit at infinity separately from the usual case, and then the "$infty$" in both of them are merely symbolic. However, we can 'add' $∞$ to the real line, to get the affinely extended real line, which is literally the closed interval $[-∞,∞]$, and the real line is literally the open interval $(-∞,∞)$.
– user21820
yesterday
1
$[-∞,∞]$ is not a field, but it is a compact topological space, and $sup(S),inf(S)$ exist for any subset $S$, and hence $limsup,liminf$ of any sequence always exists. If they are equal then the limit exists. These extend the usual ones, and may help to explain why the logical structure of the infinite limits and limits and infinity are so similar to the non-infinite case. You can check that $arctan$ on $(-∞,∞)$ can be continuously extended to $-∞$ and $∞$, and then it is literally correct to write $arctan(∞) = À/2$.
– user21820
yesterday
 |Â
show 6 more comments
up vote
7
down vote
accepted
up vote
7
down vote
accepted
It is always didactically good to give a solution for the given start. I will try it, although trigonometric substitutions from the beginning may save typing.
The substitution $t=sqrt(4-x)/x$ is ok, but we have to use improper integrals. The solution would be as follows. For $x=0+$ the corresponding (limiting) value of $t$ is $t=sqrt4/0_+=+infty$, and for $x=1$ we get $t=sqrt(4-1)/1=colorredsqrt 3$. (It is square three, maybe the one reason for writing this answer.)
$$
beginaligned
J &=
int_0^1 (1-x)sqrt4x-x^2 ; dx
\
&=
int_infty^sqrt 3
-32frac(t^2-3)t^2(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
frac(t^2+2t^2+1)-5(t^2+1)+4(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
left[
frac1(1+t^2)^2
-5frac1(1+t^2)^3
+4frac1(1+t^2)^4
right];dt
\
&=32(K_2-5K_3+4K_4) ,
endaligned
$$
where $K_n$ is the integral on the same interval of $(1+t^2)^-n$. We have the recursion
$$
beginaligned
K_n
&=int_sqrt 3^inftyt'frac1(1+t^2)^n;dt
\
&=frac t(1+t^2)^nBigg|_sqrt 3^infty
-
int_sqrt 3^infty
tcdot frac-2nt(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2nint_sqrt 3^inftyfrac(t^2+1)-1(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2n(K_n-K_n+1) ,text i.e.
\
K_n+1
&=
frac 12nleft[ (2n-1)K_n-fracsqrt34^n right] .
\[2mm]
&qquadtextThis gives:
\
K_1 &=arctan infty-arctansqrt 3 =left(frac 12-frac 13right)pi
=frac 16pi ,
\
K_2 &=frac 12left[K_1-fracsqrt 34right]=frac 112pi - fracsqrt 38 ,
\
K_3 &=frac 14left[3K_2-fracsqrt 316right]
=frac 116pi - frac7sqrt 364 ,
\
K_4 &=frac 16left[5K_3-fracsqrt 364right]
=frac 596pi - frac3sqrt 332 ,
%\
%K_5 &=frac 18left[7K_4-fracsqrt 3216right]
%=frac 35768pi - frac169sqrt 32048 ,
\
&qquadtext so putting all together
\
J&=32(K_2-5K_3+4K_4)
\
&=-frac23pi+frac 3sqrt 32 .
endaligned
$$
As said, not the quick way, but all details are displayed. Computer check:
sage: var('x,t');
sage: integral( (1-x)*sqrt(4*x-x^2), x, 0, 1 ).simplify()
-2/3*pi + 3/2*sqrt(3)
sage: integral( 32*(t^2-3)*t^2/(1+t^2)^4, t, sqrt(3), oo ).simplify()
-2/3*pi + 3/2*sqrt(3)
answered yesterday
dan_fulea
4,7631211
It is always didactically good to give a solution for the given start. I will try it, although trigonometric substitutions from the beginning may save typing.
The substitution $t=sqrt(4-x)/x$ is ok, but we have to use improper integrals. The solution would be as follows. For $x=0+$ the corresponding (limiting) value of $t$ is $t=sqrt4/0_+=+infty$, and for $x=1$ we get $t=sqrt(4-1)/1=colorredsqrt 3$. (It is square three, maybe the one reason for writing this answer.)
$$
beginaligned
J &=
int_0^1 (1-x)sqrt4x-x^2 ; dx
\
&=
int_infty^sqrt 3
-32frac(t^2-3)t^2(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
frac(t^2+2t^2+1)-5(t^2+1)+4(1+t^2)^4;dt
\
&=
+32int_sqrt 3^infty
left[
frac1(1+t^2)^2
-5frac1(1+t^2)^3
+4frac1(1+t^2)^4
right];dt
\
&=32(K_2-5K_3+4K_4) ,
endaligned
$$
where $K_n$ is the integral on the same interval of $(1+t^2)^-n$. We have the recursion
$$
beginaligned
K_n
&=int_sqrt 3^inftyt'frac1(1+t^2)^n;dt
\
&=frac t(1+t^2)^nBigg|_sqrt 3^infty
-
int_sqrt 3^infty
tcdot frac-2nt(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2nint_sqrt 3^inftyfrac(t^2+1)-1(1+t^2)^n+1;dt
\
&=-fracsqrt34^n
+
2n(K_n-K_n+1) ,text i.e.
\
K_n+1
&=
frac 12nleft[ (2n-1)K_n-fracsqrt34^n right] .
\[2mm]
&qquadtextThis gives:
\
K_1 &=arctan infty-arctansqrt 3 =left(frac 12-frac 13right)pi
=frac 16pi ,
\
K_2 &=frac 12left[K_1-fracsqrt 34right]=frac 112pi - fracsqrt 38 ,
\
K_3 &=frac 14left[3K_2-fracsqrt 316right]
=frac 116pi - frac7sqrt 364 ,
\
K_4 &=frac 16left[5K_3-fracsqrt 364right]
=frac 596pi - frac3sqrt 332 ,
%\
%K_5 &=frac 18left[7K_4-fracsqrt 3216right]
%=frac 35768pi - frac169sqrt 32048 ,
\
&qquadtext so putting all together
\
J&=32(K_2-5K_3+4K_4)
\
&=-frac23pi+frac 3sqrt 32 .
endaligned
$$
As said, not the quick way, but all details are displayed. Computer check:
sage: var('x,t');
sage: integral( (1-x)*sqrt(4*x-x^2), x, 0, 1 ).simplify()
-2/3*pi + 3/2*sqrt(3)
sage: integral( 32*(t^2-3)*t^2/(1+t^2)^4, t, sqrt(3), oo ).simplify()
-2/3*pi + 3/2*sqrt(3)
answered yesterday
dan_fulea
4,7631211
answered yesterday
dan_fulea
4,7631211
answered yesterday
dan_fulea
4,7631211
answered yesterday
dan_fulea
4,7631211
4,7631211
Are we allowed to use the infinity sign like this $arctan infty$? Because a professor of mine says that infinity sign should be used only when we have limits otherwise it's wrong to use it in other(direct) ways.
– Maths Survivor
yesterday
In this case we have the limit, explicitly $$K_1=int_sqrt 3^inftyfrac dt1+t^2=lim_Mtoinftyint_sqrt 3^Mfrac dt1+t^2=lim_MtoinftyarctanBigg|_sqrt 3^M=lim_Mtoinftyarctan M -arctansqrt 3=frac 12pi-frac 13pi .$$ To avoid this complicated notation, it is simpler to write $arctan infty$ instead of the longer $lim_Mtoinftyarctan M$. (Same in the integral.) We apply a continuous function, $arctan$, on the "symbol" $infty$, this function has a limit. The substitution was dictating the use of an improper integral, one with an $infty$ as one limit.
– dan_fulea
yesterday
I get it , thank you for your answer .
– Maths Survivor
yesterday
1
@MathsSurvivor: Your professor is correct if you are working in real analysis with the usual real line, since there is no real number $∞$. In this case, you must define infinite limit and limit at infinity separately from the usual case, and then the "$infty$" in both of them are merely symbolic. However, we can 'add' $∞$ to the real line, to get the affinely extended real line, which is literally the closed interval $[-∞,∞]$, and the real line is literally the open interval $(-∞,∞)$.
– user21820
yesterday
1
$[-∞,∞]$ is not a field, but it is a compact topological space, and $sup(S),inf(S)$ exist for any subset $S$, and hence $limsup,liminf$ of any sequence always exists. If they are equal then the limit exists. These extend the usual ones, and may help to explain why the logical structure of the infinite limits and limits and infinity are so similar to the non-infinite case. You can check that $arctan$ on $(-∞,∞)$ can be continuously extended to $-∞$ and $∞$, and then it is literally correct to write $arctan(∞) = À/2$.
– user21820
yesterday
 |Â
show 6 more comments
Are we allowed to use the infinity sign like this $arctan infty$? Because a professor of mine says that infinity sign should be used only when we have limits otherwise it's wrong to use it in other(direct) ways.
– Maths Survivor
yesterday
In this case we have the limit, explicitly $$K_1=int_sqrt 3^inftyfrac dt1+t^2=lim_Mtoinftyint_sqrt 3^Mfrac dt1+t^2=lim_MtoinftyarctanBigg|_sqrt 3^M=lim_Mtoinftyarctan M -arctansqrt 3=frac 12pi-frac 13pi .$$ To avoid this complicated notation, it is simpler to write $arctan infty$ instead of the longer $lim_Mtoinftyarctan M$. (Same in the integral.) We apply a continuous function, $arctan$, on the "symbol" $infty$, this function has a limit. The substitution was dictating the use of an improper integral, one with an $infty$ as one limit.
– dan_fulea
yesterday
I get it , thank you for your answer .
– Maths Survivor
yesterday
1
@MathsSurvivor: Your professor is correct if you are working in real analysis with the usual real line, since there is no real number $∞$. In this case, you must define infinite limit and limit at infinity separately from the usual case, and then the "$infty$" in both of them are merely symbolic. However, we can 'add' $∞$ to the real line, to get the affinely extended real line, which is literally the closed interval $[-∞,∞]$, and the real line is literally the open interval $(-∞,∞)$.
– user21820
yesterday
1
$[-∞,∞]$ is not a field, but it is a compact topological space, and $sup(S),inf(S)$ exist for any subset $S$, and hence $limsup,liminf$ of any sequence always exists. If they are equal then the limit exists. These extend the usual ones, and may help to explain why the logical structure of the infinite limits and limits and infinity are so similar to the non-infinite case. You can check that $arctan$ on $(-∞,∞)$ can be continuously extended to $-∞$ and $∞$, and then it is literally correct to write $arctan(∞) = À/2$.
– user21820
yesterday
Are we allowed to use the infinity sign like this $arctan infty$? Because a professor of mine says that infinity sign should be used only when we have limits otherwise it's wrong to use it in other(direct) ways.
– Maths Survivor
yesterday
Are we allowed to use the infinity sign like this $arctan infty$? Because a professor of mine says that infinity sign should be used only when we have limits otherwise it's wrong to use it in other(direct) ways.
– Maths Survivor
yesterday
In this case we have the limit, explicitly $$K_1=int_sqrt 3^inftyfrac dt1+t^2=lim_Mtoinftyint_sqrt 3^Mfrac dt1+t^2=lim_MtoinftyarctanBigg|_sqrt 3^M=lim_Mtoinftyarctan M -arctansqrt 3=frac 12pi-frac 13pi .$$ To avoid this complicated notation, it is simpler to write $arctan infty$ instead of the longer $lim_Mtoinftyarctan M$. (Same in the integral.) We apply a continuous function, $arctan$, on the "symbol" $infty$, this function has a limit. The substitution was dictating the use of an improper integral, one with an $infty$ as one limit.
– dan_fulea
yesterday
In this case we have the limit, explicitly $$K_1=int_sqrt 3^inftyfrac dt1+t^2=lim_Mtoinftyint_sqrt 3^Mfrac dt1+t^2=lim_MtoinftyarctanBigg|_sqrt 3^M=lim_Mtoinftyarctan M -arctansqrt 3=frac 12pi-frac 13pi .$$ To avoid this complicated notation, it is simpler to write $arctan infty$ instead of the longer $lim_Mtoinftyarctan M$. (Same in the integral.) We apply a continuous function, $arctan$, on the "symbol" $infty$, this function has a limit. The substitution was dictating the use of an improper integral, one with an $infty$ as one limit.
– dan_fulea
yesterday
I get it , thank you for your answer .
– Maths Survivor
yesterday
I get it , thank you for your answer .
– Maths Survivor
yesterday
1
1
@MathsSurvivor: Your professor is correct if you are working in real analysis with the usual real line, since there is no real number $∞$. In this case, you must define infinite limit and limit at infinity separately from the usual case, and then the "$infty$" in both of them are merely symbolic. However, we can 'add' $∞$ to the real line, to get the affinely extended real line, which is literally the closed interval $[-∞,∞]$, and the real line is literally the open interval $(-∞,∞)$.
– user21820
yesterday
@MathsSurvivor: Your professor is correct if you are working in real analysis with the usual real line, since there is no real number $∞$. In this case, you must define infinite limit and limit at infinity separately from the usual case, and then the "$infty$" in both of them are merely symbolic. However, we can 'add' $∞$ to the real line, to get the affinely extended real line, which is literally the closed interval $[-∞,∞]$, and the real line is literally the open interval $(-∞,∞)$.
– user21820
yesterday
1
1
$[-∞,∞]$ is not a field, but it is a compact topological space, and $sup(S),inf(S)$ exist for any subset $S$, and hence $limsup,liminf$ of any sequence always exists. If they are equal then the limit exists. These extend the usual ones, and may help to explain why the logical structure of the infinite limits and limits and infinity are so similar to the non-infinite case. You can check that $arctan$ on $(-∞,∞)$ can be continuously extended to $-∞$ and $∞$, and then it is literally correct to write $arctan(∞) = À/2$.
– user21820
yesterday
$[-∞,∞]$ is not a field, but it is a compact topological space, and $sup(S),inf(S)$ exist for any subset $S$, and hence $limsup,liminf$ of any sequence always exists. If they are equal then the limit exists. These extend the usual ones, and may help to explain why the logical structure of the infinite limits and limits and infinity are so similar to the non-infinite case. You can check that $arctan$ on $(-∞,∞)$ can be continuously extended to $-∞$ and $∞$, and then it is literally correct to write $arctan(∞) = À/2$.
– user21820
yesterday
 |Â
show 6 more comments
up vote
21
down vote
The us try to get rid of the square root step-by-step:
$$begineqnarray* int_0^1(1-x)sqrtxsqrt4-x,dx&stackrelxmapsto z^2=& 2int_0^1z^2(1-z^2)sqrt4-z^2,dz\&stackrelzmapsto 2u=&32int_0^1/2u^2(1-4u^2)sqrt1-u^2,du\&stackrelumapstosintheta=&32int_0^pi/6sin^2thetacos^2theta(1-4sin^2theta),dtheta\&=&4int_0^pi/6left[-1+cos(2theta)+cos(4theta)-cos(6theta)right],dtheta.endeqnarray*$$
I guess you may take it from here.
answered yesterday
Jack D'Aurizio♦
274k32268638
add a comment |Â
up vote
21
down vote
The us try to get rid of the square root step-by-step:
$$begineqnarray* int_0^1(1-x)sqrtxsqrt4-x,dx&stackrelxmapsto z^2=& 2int_0^1z^2(1-z^2)sqrt4-z^2,dz\&stackrelzmapsto 2u=&32int_0^1/2u^2(1-4u^2)sqrt1-u^2,du\&stackrelumapstosintheta=&32int_0^pi/6sin^2thetacos^2theta(1-4sin^2theta),dtheta\&=&4int_0^pi/6left[-1+cos(2theta)+cos(4theta)-cos(6theta)right],dtheta.endeqnarray*$$
I guess you may take it from here.
answered yesterday
Jack D'Aurizio♦
274k32268638
add a comment |Â
up vote
21
down vote
up vote
21
down vote
The us try to get rid of the square root step-by-step:
$$begineqnarray* int_0^1(1-x)sqrtxsqrt4-x,dx&stackrelxmapsto z^2=& 2int_0^1z^2(1-z^2)sqrt4-z^2,dz\&stackrelzmapsto 2u=&32int_0^1/2u^2(1-4u^2)sqrt1-u^2,du\&stackrelumapstosintheta=&32int_0^pi/6sin^2thetacos^2theta(1-4sin^2theta),dtheta\&=&4int_0^pi/6left[-1+cos(2theta)+cos(4theta)-cos(6theta)right],dtheta.endeqnarray*$$
I guess you may take it from here.
answered yesterday
Jack D'Aurizio♦
274k32268638
The us try to get rid of the square root step-by-step:
$$begineqnarray* int_0^1(1-x)sqrtxsqrt4-x,dx&stackrelxmapsto z^2=& 2int_0^1z^2(1-z^2)sqrt4-z^2,dz\&stackrelzmapsto 2u=&32int_0^1/2u^2(1-4u^2)sqrt1-u^2,du\&stackrelumapstosintheta=&32int_0^pi/6sin^2thetacos^2theta(1-4sin^2theta),dtheta\&=&4int_0^pi/6left[-1+cos(2theta)+cos(4theta)-cos(6theta)right],dtheta.endeqnarray*$$
I guess you may take it from here.
answered yesterday
Jack D'Aurizio♦
274k32268638
answered yesterday
Jack D'Aurizio♦
274k32268638
answered yesterday
Jack D'Aurizio♦
274k32268638
answered yesterday
Jack D'Aurizio♦
274k32268638
274k32268638
add a comment |Â
add a comment |Â
up vote
6
down vote
You can use $int_a^b f(x) dx=int_a^b f(a+b-x) dx$ to get that $$I=int_0^1 x sqrt4-(1+x)^2 dx$$ You get$$I= int_0^1 (x+1)sqrt4 - (x+1)^2 dx-int_0^1 sqrt4 - (x+1)^2 dx$$ For the first one just substitute $4-(x+1)^2 =t$ to get $$frac12 int_0^3 sqrt t dt$$and the second one is a standard square root integral.
answered yesterday
Zacky
2,1111327
How did you get $sqrt4-(1+x^2)$ shouldn't it be $sqrt4(1-x)+(1-x)^2$?
– Maths Survivor
yesterday
I used that $4x-x^2 = 4-(x-2)^2 $ to simplify faster.
– Zacky
yesterday
Also $$4(1-x)-(1-x)^2 =(1-x) (4-1+x)=(1-x) (3+x)$$$$2 ^2 - (x+1)^2 =(2-1-x) (2+1+x)=(1-x)(3+x)$$
– Zacky
yesterday
Yeah it's the same , you're right
– Maths Survivor
yesterday
1
Yes, just substitute $a+b-x=trightarrow x=a+b-t$ to prove it. The new bound will be $int_b^a$, but as we have $-dx=dt$ we can get rid of that minus sign and swap those to obtain $int_a^b f(x) dx=int_a^b f(a+b-t) dt$
– Zacky
yesterday
 |Â
show 1 more comment
up vote
6
down vote
You can use $int_a^b f(x) dx=int_a^b f(a+b-x) dx$ to get that $$I=int_0^1 x sqrt4-(1+x)^2 dx$$ You get$$I= int_0^1 (x+1)sqrt4 - (x+1)^2 dx-int_0^1 sqrt4 - (x+1)^2 dx$$ For the first one just substitute $4-(x+1)^2 =t$ to get $$frac12 int_0^3 sqrt t dt$$and the second one is a standard square root integral.
answered yesterday
Zacky
2,1111327
How did you get $sqrt4-(1+x^2)$ shouldn't it be $sqrt4(1-x)+(1-x)^2$?
– Maths Survivor
yesterday
I used that $4x-x^2 = 4-(x-2)^2 $ to simplify faster.
– Zacky
yesterday
Also $$4(1-x)-(1-x)^2 =(1-x) (4-1+x)=(1-x) (3+x)$$$$2 ^2 - (x+1)^2 =(2-1-x) (2+1+x)=(1-x)(3+x)$$
– Zacky
yesterday
Yeah it's the same , you're right
– Maths Survivor
yesterday
1
Yes, just substitute $a+b-x=trightarrow x=a+b-t$ to prove it. The new bound will be $int_b^a$, but as we have $-dx=dt$ we can get rid of that minus sign and swap those to obtain $int_a^b f(x) dx=int_a^b f(a+b-t) dt$
– Zacky
yesterday
 |Â
show 1 more comment
up vote
6
down vote
up vote
6
down vote
You can use $int_a^b f(x) dx=int_a^b f(a+b-x) dx$ to get that $$I=int_0^1 x sqrt4-(1+x)^2 dx$$ You get$$I= int_0^1 (x+1)sqrt4 - (x+1)^2 dx-int_0^1 sqrt4 - (x+1)^2 dx$$ For the first one just substitute $4-(x+1)^2 =t$ to get $$frac12 int_0^3 sqrt t dt$$and the second one is a standard square root integral.
answered yesterday
Zacky
2,1111327
You can use $int_a^b f(x) dx=int_a^b f(a+b-x) dx$ to get that $$I=int_0^1 x sqrt4-(1+x)^2 dx$$ You get$$I= int_0^1 (x+1)sqrt4 - (x+1)^2 dx-int_0^1 sqrt4 - (x+1)^2 dx$$ For the first one just substitute $4-(x+1)^2 =t$ to get $$frac12 int_0^3 sqrt t dt$$and the second one is a standard square root integral.
answered yesterday
Zacky
2,1111327
answered yesterday
Zacky
2,1111327
answered yesterday
Zacky
2,1111327
answered yesterday
Zacky
2,1111327
2,1111327
How did you get $sqrt4-(1+x^2)$ shouldn't it be $sqrt4(1-x)+(1-x)^2$?
– Maths Survivor
yesterday
I used that $4x-x^2 = 4-(x-2)^2 $ to simplify faster.
– Zacky
yesterday
Also $$4(1-x)-(1-x)^2 =(1-x) (4-1+x)=(1-x) (3+x)$$$$2 ^2 - (x+1)^2 =(2-1-x) (2+1+x)=(1-x)(3+x)$$
– Zacky
yesterday
Yeah it's the same , you're right
– Maths Survivor
yesterday
1
Yes, just substitute $a+b-x=trightarrow x=a+b-t$ to prove it. The new bound will be $int_b^a$, but as we have $-dx=dt$ we can get rid of that minus sign and swap those to obtain $int_a^b f(x) dx=int_a^b f(a+b-t) dt$
– Zacky
yesterday
 |Â
show 1 more comment
How did you get $sqrt4-(1+x^2)$ shouldn't it be $sqrt4(1-x)+(1-x)^2$?
– Maths Survivor
yesterday
I used that $4x-x^2 = 4-(x-2)^2 $ to simplify faster.
– Zacky
yesterday
Also $$4(1-x)-(1-x)^2 =(1-x) (4-1+x)=(1-x) (3+x)$$$$2 ^2 - (x+1)^2 =(2-1-x) (2+1+x)=(1-x)(3+x)$$
– Zacky
yesterday
Yeah it's the same , you're right
– Maths Survivor
yesterday
1
Yes, just substitute $a+b-x=trightarrow x=a+b-t$ to prove it. The new bound will be $int_b^a$, but as we have $-dx=dt$ we can get rid of that minus sign and swap those to obtain $int_a^b f(x) dx=int_a^b f(a+b-t) dt$
– Zacky
yesterday
How did you get $sqrt4-(1+x^2)$ shouldn't it be $sqrt4(1-x)+(1-x)^2$?
– Maths Survivor
yesterday
How did you get $sqrt4-(1+x^2)$ shouldn't it be $sqrt4(1-x)+(1-x)^2$?
– Maths Survivor
yesterday
I used that $4x-x^2 = 4-(x-2)^2 $ to simplify faster.
– Zacky
yesterday
I used that $4x-x^2 = 4-(x-2)^2 $ to simplify faster.
– Zacky
yesterday
Also $$4(1-x)-(1-x)^2 =(1-x) (4-1+x)=(1-x) (3+x)$$$$2 ^2 - (x+1)^2 =(2-1-x) (2+1+x)=(1-x)(3+x)$$
– Zacky
yesterday
Also $$4(1-x)-(1-x)^2 =(1-x) (4-1+x)=(1-x) (3+x)$$$$2 ^2 - (x+1)^2 =(2-1-x) (2+1+x)=(1-x)(3+x)$$
– Zacky
yesterday
Yeah it's the same , you're right
– Maths Survivor
yesterday
Yeah it's the same , you're right
– Maths Survivor
yesterday
1
1
Yes, just substitute $a+b-x=trightarrow x=a+b-t$ to prove it. The new bound will be $int_b^a$, but as we have $-dx=dt$ we can get rid of that minus sign and swap those to obtain $int_a^b f(x) dx=int_a^b f(a+b-t) dt$
– Zacky
yesterday
Yes, just substitute $a+b-x=trightarrow x=a+b-t$ to prove it. The new bound will be $int_b^a$, but as we have $-dx=dt$ we can get rid of that minus sign and swap those to obtain $int_a^b f(x) dx=int_a^b f(a+b-t) dt$
– Zacky
yesterday
 |Â
show 1 more comment
up vote
3
down vote
Hint:
By completing the square,
$$4x-x^2=4-(x-2)^2$$ and that hints to use the substitution
$$x-2=2cos t.$$
Then
$$int (1-x)sqrt4x-x^2dx=4int(1+2cos t)sin^2t,dt.$$
The term $sin^2t$ is integrated in the form $dfrac1-cos 2t2$, and the second term is immediate.
$$I=2t-2sin tcos t+dfrac83sin^3t=\arccosdfracx-22-(x-2)sqrt1-dfrac(x-2)^24+dfrac83left(1-dfrac(x-2)^24right)^3/2$$
answered yesterday
Yves Daoust
114k665209
add a comment |Â
up vote
3
down vote
Hint:
By completing the square,
$$4x-x^2=4-(x-2)^2$$ and that hints to use the substitution
$$x-2=2cos t.$$
Then
$$int (1-x)sqrt4x-x^2dx=4int(1+2cos t)sin^2t,dt.$$
The term $sin^2t$ is integrated in the form $dfrac1-cos 2t2$, and the second term is immediate.
$$I=2t-2sin tcos t+dfrac83sin^3t=\arccosdfracx-22-(x-2)sqrt1-dfrac(x-2)^24+dfrac83left(1-dfrac(x-2)^24right)^3/2$$
answered yesterday
Yves Daoust
114k665209
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hint:
By completing the square,
$$4x-x^2=4-(x-2)^2$$ and that hints to use the substitution
$$x-2=2cos t.$$
Then
$$int (1-x)sqrt4x-x^2dx=4int(1+2cos t)sin^2t,dt.$$
The term $sin^2t$ is integrated in the form $dfrac1-cos 2t2$, and the second term is immediate.
$$I=2t-2sin tcos t+dfrac83sin^3t=\arccosdfracx-22-(x-2)sqrt1-dfrac(x-2)^24+dfrac83left(1-dfrac(x-2)^24right)^3/2$$
answered yesterday
Yves Daoust
114k665209
Hint:
By completing the square,
$$4x-x^2=4-(x-2)^2$$ and that hints to use the substitution
$$x-2=2cos t.$$
Then
$$int (1-x)sqrt4x-x^2dx=4int(1+2cos t)sin^2t,dt.$$
The term $sin^2t$ is integrated in the form $dfrac1-cos 2t2$, and the second term is immediate.
$$I=2t-2sin tcos t+dfrac83sin^3t=\arccosdfracx-22-(x-2)sqrt1-dfrac(x-2)^24+dfrac83left(1-dfrac(x-2)^24right)^3/2$$
answered yesterday
Yves Daoust
114k665209
edited 20 hours ago
answered yesterday
Yves Daoust
114k665209
answered yesterday
Yves Daoust
114k665209
answered yesterday
Yves Daoust
114k665209
114k665209
add a comment |Â
add a comment |Â
up vote
3
down vote
Your approach can be completed by noting that the domain of integrtion $xin(0,1)$ gets transformed to the unbounded domain of integration $tin(sqrt3,infty)$ with reversal of orientation (which negates the integral).
We can also set $u=1-x/2implies4!left(1-u^2right)=4x-x^2$. Therefore,
$$
int_0^1(1-x)sqrt4x-x^2,mathrmdx=4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
$$
Then, setting $u=sin(theta)$ gives
$$
beginalign
4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
&=4int_pi/6^pi/2(2sin(theta)-1)overbracecos^2(theta)^1-sin^2(theta),mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00sin^3(theta)+color#090sin^2(theta)+2sin(theta)-1right)mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00frac3sin(theta)-sin(3theta)4+color#090frac1-cos(2theta)2+2sin(theta)-1right)mathrmdtheta\
%&=-left[frac23cos(3theta)+sin(2theta)+2cos(theta)+2thetaright]_pi/6^pi/2\[3pt]
%&=frac32sqrt3-frac23pi
endalign
$$
answered yesterday
robjohn♦
259k26298613
At the stage $(2s-1)c^2$, it is more direct to integrate $sc^2$ and $c^2$.
– Yves Daoust
20 hours ago
The mechanics should include $sc^n$ and $cs^n$, as well as $s^2m+1c^n$ and $c^2m+1s^n$ where you expand the even powers. Polynomials are easy.
– Yves Daoust
18 hours ago
You needn't tell me.
– Yves Daoust
18 hours ago
add a comment |Â
up vote
3
down vote
Your approach can be completed by noting that the domain of integrtion $xin(0,1)$ gets transformed to the unbounded domain of integration $tin(sqrt3,infty)$ with reversal of orientation (which negates the integral).
We can also set $u=1-x/2implies4!left(1-u^2right)=4x-x^2$. Therefore,
$$
int_0^1(1-x)sqrt4x-x^2,mathrmdx=4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
$$
Then, setting $u=sin(theta)$ gives
$$
beginalign
4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
&=4int_pi/6^pi/2(2sin(theta)-1)overbracecos^2(theta)^1-sin^2(theta),mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00sin^3(theta)+color#090sin^2(theta)+2sin(theta)-1right)mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00frac3sin(theta)-sin(3theta)4+color#090frac1-cos(2theta)2+2sin(theta)-1right)mathrmdtheta\
%&=-left[frac23cos(3theta)+sin(2theta)+2cos(theta)+2thetaright]_pi/6^pi/2\[3pt]
%&=frac32sqrt3-frac23pi
endalign
$$
answered yesterday
robjohn♦
259k26298613
At the stage $(2s-1)c^2$, it is more direct to integrate $sc^2$ and $c^2$.
– Yves Daoust
20 hours ago
The mechanics should include $sc^n$ and $cs^n$, as well as $s^2m+1c^n$ and $c^2m+1s^n$ where you expand the even powers. Polynomials are easy.
– Yves Daoust
18 hours ago
You needn't tell me.
– Yves Daoust
18 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Your approach can be completed by noting that the domain of integrtion $xin(0,1)$ gets transformed to the unbounded domain of integration $tin(sqrt3,infty)$ with reversal of orientation (which negates the integral).
We can also set $u=1-x/2implies4!left(1-u^2right)=4x-x^2$. Therefore,
$$
int_0^1(1-x)sqrt4x-x^2,mathrmdx=4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
$$
Then, setting $u=sin(theta)$ gives
$$
beginalign
4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
&=4int_pi/6^pi/2(2sin(theta)-1)overbracecos^2(theta)^1-sin^2(theta),mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00sin^3(theta)+color#090sin^2(theta)+2sin(theta)-1right)mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00frac3sin(theta)-sin(3theta)4+color#090frac1-cos(2theta)2+2sin(theta)-1right)mathrmdtheta\
%&=-left[frac23cos(3theta)+sin(2theta)+2cos(theta)+2thetaright]_pi/6^pi/2\[3pt]
%&=frac32sqrt3-frac23pi
endalign
$$
answered yesterday
robjohn♦
259k26298613
Your approach can be completed by noting that the domain of integrtion $xin(0,1)$ gets transformed to the unbounded domain of integration $tin(sqrt3,infty)$ with reversal of orientation (which negates the integral).
We can also set $u=1-x/2implies4!left(1-u^2right)=4x-x^2$. Therefore,
$$
int_0^1(1-x)sqrt4x-x^2,mathrmdx=4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
$$
Then, setting $u=sin(theta)$ gives
$$
beginalign
4int_1/2^1(2u-1)sqrt1-u^2,mathrmdu
&=4int_pi/6^pi/2(2sin(theta)-1)overbracecos^2(theta)^1-sin^2(theta),mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00sin^3(theta)+color#090sin^2(theta)+2sin(theta)-1right)mathrmdtheta\
&=4int_pi/6^pi/2left(-2color#C00frac3sin(theta)-sin(3theta)4+color#090frac1-cos(2theta)2+2sin(theta)-1right)mathrmdtheta\
%&=-left[frac23cos(3theta)+sin(2theta)+2cos(theta)+2thetaright]_pi/6^pi/2\[3pt]
%&=frac32sqrt3-frac23pi
endalign
$$
answered yesterday
robjohn♦
259k26298613
edited 18 hours ago
answered yesterday
robjohn♦
259k26298613
answered yesterday
robjohn♦
259k26298613
answered yesterday
robjohn♦
259k26298613
259k26298613
At the stage $(2s-1)c^2$, it is more direct to integrate $sc^2$ and $c^2$.
– Yves Daoust
20 hours ago
The mechanics should include $sc^n$ and $cs^n$, as well as $s^2m+1c^n$ and $c^2m+1s^n$ where you expand the even powers. Polynomials are easy.
– Yves Daoust
18 hours ago
You needn't tell me.
– Yves Daoust
18 hours ago
add a comment |Â
At the stage $(2s-1)c^2$, it is more direct to integrate $sc^2$ and $c^2$.
– Yves Daoust
20 hours ago
The mechanics should include $sc^n$ and $cs^n$, as well as $s^2m+1c^n$ and $c^2m+1s^n$ where you expand the even powers. Polynomials are easy.
– Yves Daoust
18 hours ago
You needn't tell me.
– Yves Daoust
18 hours ago
At the stage $(2s-1)c^2$, it is more direct to integrate $sc^2$ and $c^2$.
– Yves Daoust
20 hours ago
At the stage $(2s-1)c^2$, it is more direct to integrate $sc^2$ and $c^2$.
– Yves Daoust
20 hours ago
The mechanics should include $sc^n$ and $cs^n$, as well as $s^2m+1c^n$ and $c^2m+1s^n$ where you expand the even powers. Polynomials are easy.
– Yves Daoust
18 hours ago
The mechanics should include $sc^n$ and $cs^n$, as well as $s^2m+1c^n$ and $c^2m+1s^n$ where you expand the even powers. Polynomials are easy.
– Yves Daoust
18 hours ago
You needn't tell me.
– Yves Daoust
18 hours ago
You needn't tell me.
– Yves Daoust
18 hours ago
add a comment |Â
up vote
2
down vote
Hint: $(1-x)sqrt4x-x^2 dx=-sqrt4x-x^2dx +(2-x)sqrt4x-x^2dx=-sqrt4x-x^2dx +0.5sqrt4x-x^2d(4x-x^2)$
answered yesterday
Vasya
2,6531514
add a comment |Â
up vote
2
down vote
Hint: $(1-x)sqrt4x-x^2 dx=-sqrt4x-x^2dx +(2-x)sqrt4x-x^2dx=-sqrt4x-x^2dx +0.5sqrt4x-x^2d(4x-x^2)$
answered yesterday
Vasya
2,6531514
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: $(1-x)sqrt4x-x^2 dx=-sqrt4x-x^2dx +(2-x)sqrt4x-x^2dx=-sqrt4x-x^2dx +0.5sqrt4x-x^2d(4x-x^2)$
answered yesterday
Vasya
2,6531514
Hint: $(1-x)sqrt4x-x^2 dx=-sqrt4x-x^2dx +(2-x)sqrt4x-x^2dx=-sqrt4x-x^2dx +0.5sqrt4x-x^2d(4x-x^2)$
answered yesterday
Vasya
2,6531514
answered yesterday
Vasya
2,6531514
answered yesterday
Vasya
2,6531514
answered yesterday
Vasya
2,6531514
2,6531514
add a comment |Â
add a comment |Â
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5
I'd write $4x-x^2=4-(x-2)^2$ and write $x-2=2sintheta$.
– Lord Shark the Unknown
yesterday
the domain of integration $xin(0,1)$ should be transformed into $tin(3,infty)$ with a reversal of orientation.
– robjohn♦
yesterday