Parameter-dependent integral $\int_0^1\sin(a+\arcsin(x))dx$ I am calculating this integral $I(a)=\int_0^1\sin(a+\arcsin(x))dx$. I was given the hint to use rules for calculating Parameter-dependent integrals, but I don't really know how to proceed.
My first guess was to use addition theorem:
$$
\begin{align} I(a) &= \int_0^1 \sin(a+\arcsin(x))dx
\\ &=\int_0^1 \sin(a)\cos(\arcsin(x))dx + \int_0^1 \cos(a)\sin(\arcsin(x))dx
\\ &=-\sin(a)\int_0^1 \cos(\arcsin(x))dx+\frac{\cos(a)}{2}
\end{align}
$$
I also calculated the derivative $I'(a)$ and get:
$$
I'(a)=-\int_0^1 \cos(a+\arcsin(x))dx=-\cos(a)\int_0^1 \cos(\arcsin(x))dx+\frac{\sin(a)}{2}
$$
I tried to expose $\int_0^1\cos(\arcsin(x))dx$ and get:
$$
I(a)\cos(a)=-I'(a)\sin(a)+\frac{1}{2}
$$
But after that I really don't know how to proceed.

Note that from
$$I(a)\cos a+I'(a)\sin a=\frac12$$
we have
$$[I(a)\sin a]'=\frac{1}{2}$$
thus $$I(a)\sin a=\frac{a}{2}+c$$
or $$I(a)=(\frac{a}{2}+c)\csc a$$
