How can I find the integral of this function using trig substitution? $$\int_0^1x^3\sqrt{1 - x^2}dx$$
I need to find the integral of this function using trigonometric substitution. 
Using triangles, I found that $x = \sin\theta$, and $dx = \cos\theta d\theta$; so I have
$$\int_0^{\pi/2}\sin^3\theta\sqrt{1 - \sin^2\theta}\cos\theta d\theta$$ then, using identities:
$$\int_0^{\pi/2}\sin^3\theta\sqrt{\cos^2\theta}\cos\theta d\theta$$
$$\int_0^{\pi/2}\sin^3\theta\cos^2\theta d\theta$$
After this point, I don't know where to go. My teacher posted solutions, but I don't quite understand it.
Does anyone know how this can be solved using trig substitution? The answer is $2/15$. 
Much thanks,
Zolani13
 A: Factor out a $\sin\theta$ from the $\sin^3\theta$. Write the remaining $\sin^2\theta$ as $1-\cos^2\theta$. Now let $u=\cos\theta$.
$$
\sin^3\theta\, \cos^2\theta  \,d\theta= (1- {\cos^2\theta} )\cos^2\theta\sin\theta \,d\theta
= (\cos^2\theta- {\cos^4\theta} )  \sin\theta \,d\theta\ \ 
\buildrel{u=\cos\theta}\over = \ \ -(u^2-u^4)\,du
$$
A: A couple of comments
1) if $x = \sin \theta$ then you write $dx = \cos \theta d\theta$. (You missed the $d\theta$).
2) Since you are integrating in the range $0$, $\pi/2$, $\sqrt{\cos^2 \theta} = \cos \theta$. You should probably mention that in your work.
Now you are at the integral
$$\int_{0}^{\pi/2} \sin^3 \theta \cos ^2 \theta d \theta$$
Normally when you have to integrate something like $\sin^{2n+1}\theta \cos^m \theta$, the substitution $t = \cos \theta$ usually works, because the $dt$ swallows one of the $\sin \theta$, and then you can use $\sin^{2n} \theta = (1 - \cos^2 \theta)^n$
A: From here, convert $\sin^3(\theta)\cos^2(\theta) = \sin(\theta)(1 - \cos^2(\theta))\cos^2(\theta) = \sin(\theta)\cos^2(\theta) - \sin(\theta)\cos^4(\theta)$.  Now, solve both integrals using a u-substitution: $u = \cos(\theta), du = -\sin(\theta) d\theta $. Can you finish from here?
