Set u=1-x^2u=1−x2. Then du=-2xdu=−2x We don't have a -2−2 though. So we'll multiply by -2/-2−2−2. Put the -2−2 on the top of the numerator inside the integral and leave the 1/-21−2 outside the integral.
1/-2 int_0^1 -2xsqrt(1-x^2)dx1−2∫10−2x√1−x2dx
Make the substitution. But note that if you substitute, you have to change the integrand:
For x=0x=0: u=1-0^2=1u=1−02=1
For x=1x=1: u=1-1^2=0u=1−12=0
So we now have
-1/2 int_1^0 sqrt(u) du−12∫01√udu
We can flip the integrand by using the negative on the outside. We can also write sqrt(u)√u as u^(1/2)u12 so it's easier to integrate:
1/2 int_0^1 u^(1/2) du12∫10u12du
Now integrate:
1/2 (u^(3/2))/(3/2)12u3232
Take out the 1/3/213/2 and then plug in the integrands:
(1/2)(1/(3/2)) (1^(3/2)-0^(3/2))(12)(132)(132−032)
Simplify:
(1/3)(1)=1/3(13)(1)=13
