Solving integral $\int \arcsin x \cos x dx$ Can anyone give me a hint how to solve $$\int \arcsin(x)\cos(x)dx ~?$$ 
 A: Although the integrand does not possess an elementary anti-derivative, we can nevertheless express its definite counterpart in terms of special Struve functions, as $\dfrac\pi2\bigg[\sin(1)-H_0(1)\bigg]$.
A: $\int\sin^{-1}x\cos x~dx$
$=\int\sin^{-1}x~d(\sin x)$
$=\sin^{-1}x\sin x-\int\sin x~d(\sin^{-1}x)$
$=\sin^{-1}x\sin x-\int\dfrac{\sin x}{\sqrt{1-x^2}}dx$
Let $x=\sin u$ ,
Then $dx=\cos u~du$
$\therefore\sin^{-1}x\sin x-\int\dfrac{\sin x}{\sqrt{1-x^2}}dx$
$=\sin^{-1}x\sin x-\int\sin\sin u~du$
$=\sin^{-1}x\sin x-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}u}{(2n+1)!}du$
$=\sin^{-1}x\sin x+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}u}{(2n+1)!}d(\cos u)$
$=\sin^{-1}x\sin x+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\cos^2u)^n}{(2n+1)!}d(\cos u)$
$=\sin^{-1}x\sin x+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^k\cos^{2k}u}{(2n+1)!}d(\cos u)$
$=\sin^{-1}x\sin x+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\cos^{2k}u}{(2n+1)!k!(n-k)!}d(\cos u)$
$=\sin^{-1}x\sin x+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\cos^{2k+1}u}{(2n+1)!k!(n-k)!(2k+1)}+C$
$=\sin^{-1}x\sin x+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!(1-x^2)^{k+\frac{1}{2}}}{(2n+1)!k!(n-k)!(2k+1)}+C$
