How to compute $\int_0^a \sin(\tan^{-1}(b\sin\theta))\ d\theta$ How to compute
$$\int_0^a \sin(\tan^{-1}(b\sin\theta))\ d\theta$$
I've tried to rewrite it as
$$\int_0^a\frac{b\sin\theta}{\sqrt{b^2\sin^2\theta+1}}d\theta$$
But I'm still stuck here.
 A: Here one way to go about just using the regular techniques one may learn in a calc class.
The way you have it written: $$b\int_{0}^{a}\frac{\sin\theta}{\sqrt{b^{2}\sin^{2}\theta +1}}d\theta$$ is a good start.
Now, use the identity $\sin\theta = 1-\cos^{2}\theta$
$$b\int_{0}^{a}\frac{\sin\theta}{\sqrt{b^{2}(1-\cos^{2}\theta)+1}}d\theta$$
Let $u=\cos\theta, \;\ du=-\sin\theta d\theta$
$$-b\int\frac{1}{\sqrt{b^{2}(1-u^{2})}}du$$
Let $u=\sin(y)\frac{\sqrt{b^{2}+1}}{b}, \;\ du=\frac{\sqrt{b^{2}+1}}{b}\cos(y), \;\ y=\sin^{-1}\left(\frac{ub}{\sqrt{b^{2}+1}}\right)$
Making these subs now results in everything cancelling down to:
$$-b\int \frac{\sqrt{b^{2}+1}}{b}\cos(y)\cdot \frac{1}{\cos(y)\sqrt{b^{2}+1}}dy$$
$$=-\int dy$$
$$y=-\sin^{-1}\left(\frac{ub}{\sqrt{b^{2}+1}}\right)$$
Revert $u=\cos(\theta)$
$$-\sin^{-1}\left(\frac{b\cos(\theta)}{\sqrt{b^{2}+1}}\right)$$
Now, use the limits of integration to get:
$$\tan^{-1}(b)-\sin^{-1}\left(\frac{b\cos(a)}{\sqrt{b^{2}+1}}\right)$$
A: The change of variables
$$
\cos\theta=\frac{b+1}{b}x
$$
gives
$$
\int\frac{b\sin\theta}{\sqrt{b^2\sin^2\theta+1}}d\theta=-\sqrt{b^2+1}\int\frac{dx}{\sqrt{1-x^2}}.
$$
Let's check it:
$$
\sin\theta\,d\theta=-\frac{b+1}{b}dx,
$$
$$
b^2\sin^2\theta+1=b^2(1-\cos^2\theta)+1=b^2(1-\frac{(b+1)^2}{b^2}x^2)+1=(b^2+1)(1-x^2).
$$
