Calculate $\int\frac{a}{\sin(x)\sqrt{\sin^2(x)-a^2}}dx$

Question

We have $a \in \mathbb{R}$. I tried by using substitution $u=\cot(x)$. This gives $du=-\frac{1}{\sin^2(x)}dx$.

So $$\int\frac{a}{\sin(x)\sqrt{\sin^2(x)-a^2}}dx=\int\frac{-a\sin^2(x)}{\sin(x)\sqrt{\sin^2(x)-a^2}}du=\int\frac{-a\sin(x)}{\sqrt{\sin^2(x)-a^2}}du.$$

Using $u=\frac{\cos(x)}{\sin(x)}$, we get $$\int\frac{-a\cos(x)}{\frac{u}{u}\sqrt{\cos^2(x)-a^2u^2}}du=-\int\frac{-a\cos(x)}{\sqrt{\cos^2(x)-a^2u^2}}du.$$

At this point I'm lost - perhaps this was the wrong substitution to make?

Of course, $\int\frac{-a}{\sqrt{b^2-a^2u^2}}du=\arccos(\frac{au}{b})+\text{const}.$

But I'm not sure if this helps since $\cos(x)$ depends on $u$ (i.e $x=\cot^{-1}(u)$)?

Answer 1

hint

the integrand will be written as

$$\frac{a\sin(x)dx}{\sin^2(x)\sqrt{\sin^2(x)-a^2}}=$$

$$\frac{a\sin(x)dx}{(1-\cos^2(x))\sqrt{1-\cos^2(x)-a^2}}$$

put $$t=\cos(x)$$

Answer 2

Hint: We have, taking out $\sin x$ from radical, $$I=a \int \frac {\csc^2 x dx}{\sqrt {(1-a^2)-a^2\cot^2 x}}$$ Now put $t=\cot x$.

Can you finish?