closed-form of an integral with regard to $a$

Question

Suppose that $a$ is a constant and $a>1$. So how can we evaluate the integral $$ I(a) = \int_0^1 \frac{t}{(a-t)\sqrt{1-t^2}}\;dt $$ I just wonder if there is a closed-form. Thank you.

Answer 1

It can be transformed in a trigonometric integral using the change $t = \sin z$: $$\int\frac{t}{(a-t)\sqrt{1-t^2}}\,dt = \int\frac{a\sin z}{a-\sin z}\,dz.$$ Now you can use the Weierstrass substitution.

The form of the solution depends of $|a|$ ($<1$, $=1$, $>1$).

Answer 2

I would suggest adding and subtracting $a$ in the numerator, $$ \frac{t}{(a-t)\sqrt{1-t^2}}=\frac{t-a+a}{(a-t)\sqrt{1-t^2}}=-\frac{1}{\sqrt{1-t^2}}+\frac{a}{(a-t)\sqrt{1-t^2}}. $$ The first term gives an inverse sine, for the second term, let $$ u=\frac{at-1}{\sqrt{1-t^2}} $$ and you will get a simple primitive. Since you did not show any effort, I will stop here.