How to integrate this irrational function?

Question

Evaluate

$$ \int \dfrac{\sqrt{1-x^2}}{ax+b} \ dx $$

where $a, b \in \mathbb{R}$

I don't know how to solve this.

Please help.
Thanks in advance.

Answer 1

Substitute $$x=\sin(u),\mathrm dx=\cos(u)\mathrm du$$ then you will get $$\int\frac{\cos^2(u)}{a\sin(u)+b}\mathrm du$$ and now use $$s=\tan\left(\frac{u}{2}\right)$$ and $$\mathrm ds=\frac{1}{2}\sec^2\left(\frac{u}{2}\right)\mathrm du$$

Answer 2

As I suggested in a comment:

$$x=\sin t$$

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$$\int \dfrac{\sqrt{1-x^2}}{ax+b} \ dx=\int \frac{\cos^2 t}{a \sin t+b}dt=$$

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$$\tan (t/2)=u$$

$$\sin t=\frac{2u}{1+u^2}$$

$$\cos t=\frac{1-u^2}{1+u^2}$$

$$dt=\frac{2du}{1+u^2}$$

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$$=2 \int \frac{(1-u^2)^2}{(1+u^2)^3 \left(\frac{2 a u}{1+u^2}+b \right)}du=\frac{2}{b} \int \frac{(1-u^2)^2}{(1+u^2)^2 \left(u^2+2\frac{a}{b}u+1 \right)}du$$

Now we could use partial fractions on the resulting rational function.

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Euler substitution on the initial expression could lead to something better, but I haven't tried it.