Integration involving $\arcsin$ How to integrate the following function:
$$\int_0^a \arcsin\sqrt{a \over a+x}dx$$
By using the substitution $x = a\tan^2\theta$, I managed to write the integral as:
$$2a\int_0^{\pi \over 4}\theta \frac{\sin\theta}{\cos^3\theta}d\theta$$
How would I proceed? Should I use by parts method?
 A: A different way, maybe with easier calculations:
As $x$ runs from $0$ to $a$ we find that $y=f(x)=\arcsin\sqrt{a/(a+x)}$ decreases monotonically from $\pi/2$ to $\pi/4$. Thus (draw a figure)
$$
\int_0^a \arcsin\sqrt{\frac{a}{a+x}}\,dx=\int_{\pi/4}^{\pi/2}f^{-1}(y)\,dy+a\times\frac{\pi}{4}.
$$
But
$$
f^{-1}(y)=a\cot^2y,
$$
so (taking the constant part into the integral)
$$
\int_0^a \arcsin\sqrt{\frac{a}{a+x}}\,dx=a\int_{\pi/4}^{\pi/2}\cot^2y+1\,dy
=a\bigl[-\cot y\bigr]_{\pi/4}^{\pi/2}=a.
$$
A: 
$$\int \arcsin\sqrt{a \over a+x}dx$$

Set $t=x+a$ and $dt=dx$
$$\int \arcsin \left(\sqrt a \sqrt{\frac 1 t}\right)dt$$
Now by parts $f=\arcsin\left(\sqrt a \sqrt{\frac 1 t}\right)$ and $g=t$
$$=t \arcsin\left(\sqrt a\sqrt{\frac 1 t}\right)+\frac 1 2 \int \frac{\sqrt{\frac 1 t}}{\sqrt{\frac 1 a-\frac 1 t}}$$
$$=t \arcsin \left(\sqrt a \sqrt{\frac 1 t}\right)+\frac{\sqrt a}{2}\int\frac{dt}{\sqrt{t-a}}$$
$$=\sqrt a \sqrt{t-a}+t\arcsin\left(\sqrt a \sqrt{\frac 1 t}\right)+\mathcal C$$
$$=\sqrt a \sqrt{x}+(x+a)\arcsin\left(\sqrt a \sqrt{\frac{1}{x+a}}\right)+\mathcal C$$
$$\left(\sqrt a \sqrt{x}+(x+a)\arcsin\left(\sqrt a \sqrt{\frac{1}{x+a}}\right)\right)\bigg|_0^a=a$$
A: Integrating by parts, $$\int\arcsin\sqrt{\dfrac a{x+a}}\ dx$$
$$=\arcsin\sqrt{\dfrac a{x+a}}\int\ dx-\int\left(\dfrac{d(\arcsin\sqrt{\dfrac a{x+a}})}{dx}\int\ dx\right)dx$$
$$=x\arcsin\sqrt{\dfrac a{x+a}}-\dfrac{\sqrt a}2\int\left(\dfrac{x+a}x\cdot\dfrac1{(x+a)^{3/2}}\cdot x\right)dx$$
$$=x\arcsin\sqrt{\dfrac a{x+a}}-\dfrac{\sqrt a}2\int\dfrac{dx}{\sqrt{x+a}}=?$$
A: You are on the right track.
By parts,
$$\int\theta \frac{\sin\theta}{\cos^3\theta}d\theta=\frac\theta{2\cos^2(\theta)}-\int\frac{d\theta}{2\cos^2(\theta)}=\frac\theta{2\cos^2(\theta)}-\frac{\tan(\theta)}2.$$
A: HINT(OTHER APPROACH)
Let  $u=a+x$, then we have
\begin{align}
\int{arcsin\sqrt{\frac{x-u}{u}}}du
\end{align}
Using Integration By Parts\, let $f=arcsin\sqrt{\frac{x-u}{u}}du$ and $dg=du$
We obtain
\begin{align}
=x arcsin(\sqrt{\frac{a}{a+x}})+\frac{{\sqrt{\frac{ax}{(a+x)^2}}(a+x)(\sqrt x-\sqrt a\arctan(\sqrt{\frac{\sqrt x}{\sqrt a}})}}{\sqrt x}+C
\end{align}
