Problem with the change of variable theorem of integrals

[IvanT]

Homework Statement

So I came across the integral \int^{1}_{0}x\sqrt{1-x^2} and I tried to solve it in two ways using the change of variables theorem for integration, however both ways are supposed to give me the same result, but they differ in the sign and I cannot find what I am doing wrong.

Homework Equations

The change of variable theorem states the following: Let I=[a,b]. Let g: I->R be a function of class C^1, with g'(x) different from 0 for x in (a,b). Then the set g(I) is a closed interval J with end points g(a) and g(b) (not necessarily in that order). If f: J->R is continous, then \int^{g(b)}_{g(a)}f=\int^{b}_{a}(f°g)g'

The Attempt at a Solution

So, for the first solution I used g(x)=cos(x) in the interval I=[0,Pi/2]. Clearly cos(I)=[0,1] and g'(x)=-sin(x) is not 0 for any x in (0,pi/2). Since f(x)=x*sqrt(1-x^2) is continuous in [0,1] the hypotheses of the theorem are satisfied and we have \int^{g(pi/2)}_{g(0)}f=\int^{0}_{1}x\sqrt{1-x^2}=-\int^{1}_{0}x\sqrt{1-x^2}=\int^{pi/2}_{0}cos(x)sin(x)(-sin(x))=-sin^3(pi/2)/3+sin^3(0)/3=-1/3 so that \int^{1}_{0}f=1/3.
On the other hand, if I consider the same g(x)=cos(x) but now in the interval I=[-pi/2,0], we have again that cos(I)=[0,1] and the hypotheses are satisfied. However we get \int^{g(0)}_{g(-pi/2)}f=\int^{1}_{0}x\sqrt{1-x^2}=\int^{0}_{-pi/2}cos(x)sin(x)(-sin(x))=-sin^3(0)/3+sin^3(-pi/2)/3=-1/3 so that \int^{1}_{0}f=-1/3!.
I checked every step a few times already and I cannot find the mistake. Any help would be appreciated.

 

[algebrat]

Homework Statement

So I came across the integral \int^{1}_{0}x\sqrt{1-x^2} and I tried to solve it in two ways using the change of variables theorem for integration, however both ways are supposed to give me the same result, but they differ in the sign and I cannot find what I am doing wrong.

Homework Equations

The change of variable theorem states the following: Let I=[a,b]. Let g: I->R be a function of class C^1, with g'(x) different from 0 for x in (a,b). Then the set g(I) is a closed interval J with end points g(a) and g(b) (not necessarily in that order). If f: J->R is continous, then \int^{g(b)}_{g(a)}f=\int^{b}_{a}(f°g)g'

The Attempt at a Solution

So, for the first solution I used g(x)=cos(x) in the interval I=[0,Pi/2]. Clearly cos(I)=[0,1] and g'(x)=-sin(x) is not 0 for any x in (0,pi/2). Since f(x)=x*sqrt(1-x^2) is continuous in [0,1] the hypotheses of the theorem are satisfied and we have \int^{g(pi/2)}_{g(0)}f=\int^{0}_{1}x\sqrt{1-x^2}=-\int^{1}_{0}x\sqrt{1-x^2}=\int^{pi/2}_{0}cos(x)sin(x)(-sin(x))=-sin^3(pi/2)/3+sin^3(0)/3=-1/3 so that \int^{1}_{0}f=1/3.
On the other hand, if I consider the same g(x)=cos(x) but now in the interval I=[-pi/2,0], we have again that cos(I)=[0,1] and the hypotheses are satisfied. However we get \int^{g(0)}_{g(-pi/2)}f=\int^{1}_{0}x\sqrt{1-x^2}=\int^{0}_{-pi/2}cos(x)sin(x)(-sin(x))=-sin^3(0)/3+sin^3(-pi/2)/3=-1/3 so that \int^{1}_{0}f=-1/3!.
I checked every step a few times already and I cannot find the mistake. Any help would be appreciated.

When you took √sin²(x), it's |sin(x)|, which in the second region is -sin(x)

 

[IvanT]

When you took √sin²(x), it's |sin(x)|, which in the second region is -sin(x)

Oh xD, I didn't noticed that, thanks a lot : ).