# Homework Help: Problem with the change of variable theorem of integrals

1. Jun 20, 2012

### IvanT

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 $\int^{g(b)}_{g(a)}f$=$\int^{b}_{a}(f°g)g'$

3. 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 continous 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.

Last edited: Jun 20, 2012
2. Jun 20, 2012

### algebrat

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

3. Jun 20, 2012

### IvanT

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