Integrating $x^5 \arcsin x$ Integrating $$\int_0^1 x^5 (\sin^{-1}x) \, dx$$
the answer is $\dfrac{11 \pi}{192}$.
I did substitute $x=\sin^2 u$ and obtained $$2\int_0^{\pi/2}( \sin^{12}x)( \cos x) \, dx$$ but got the incorrect answer. Why isn't this substitution working?
$$\int_0^{\pi/2}( \sin^m x)( \cos^n x)=\frac{(m-1)(m-3)\cdots(1\text{ or } 2)(n-1)(n-3) \cdots (1\text{ or } 2)k}{(m+n)(m+n-2)(m+n-4)\cdots(1\text{ or } 2)}$$ $k=\frac{\pi}{2}$ if $m$ and $n$ are even else $1$.
 A: I think you were on the right track of using a trig sub but i would use $x = \sin u$
$$
\int \left(\sin^5 u \right)u \cos u du = \frac{1}{6}\int u \dfrac{d}{du}\sin^6 u du
$$
use by parts
A: We may use the substitution $x=\sin(\theta)$ followed by integration by parts to get:
$$\begin{eqnarray*} I = \int_{0}^{\pi/2}\theta\cos(\theta)\sin(\theta)^5\,d\theta &=& \frac{\pi}{12}-\frac{1}{6}\int_{0}^{\pi/2}\sin(\theta)^6\,d\theta\\&=&\frac{\pi}{12}-\frac{1}{6}\int_{0}^{\pi/2}\cos(\theta)^6\,d\theta\tag{1}\end{eqnarray*}$$
By exploiting the formula $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and the binomial theorem we have:
$$ \cos^6(\theta) = \frac{5}{16}+\frac{15}{32}\cos(2\theta)+\frac{3}{16}\cos(4\theta)+\frac{1}{32}\cos(6\theta)\tag{2} $$
hence:
$$ I = \frac{\pi}{12}-\frac{1}{6}\cdot\frac{\pi}{2}\cdot \frac{5}{16}=\color{red}{\frac{11\pi}{192}}\tag{3}$$
follows.
A: $$
x = (\sin u)^2
$$
$$
\sin^{-1} x = \sin^{-1}((\sin u)^2)
$$
If you had $\sin^{-1}(\sin u)$, then (assuming $-\pi/2\le u\le\pi/2$), that would simplify to $u$.  But that is not what you have.
