<s>[INST] How to solve the integral $\int\limits_0^a {\frac{{\sqrt {{a^2} - {x^2}} }}{{b - x}}} \mathop{\mathrm{d}x}\\$? I have seen this integral:
$$\int\limits_0^a {\frac{{\sqrt {{a^2} - {x^2}} }}{{b - x}}} \mathop{\mathrm{d}x}\\$$
In this integral: $a$ and $b$ are constants.

I have try with two ways, but failed:


*

*$u = \sqrt {{a^2} - {x^2}}$

*$x = a\sin t$


It seems that they are not true-way to solve this integral.
Any suggestion for solving this integral?


Now, I am not a student; so, this is not my exercise.
  Sorry about my English. If my question is not clear, please comment below this question.

 [/INST] First off I would let $x=-a\cos\theta$. Then
$$\begin{align}\int_0^a\frac{\sqrt{a^2-x^2}}{b-x}dx&=\int_{\frac{\pi}2}^{\pi}\frac{a^2\sin^2\theta}{b+a\cos\theta}d\theta\\
&=\int_{\frac{\pi}2}^{\pi}\frac{a^2(1-\cos^2\theta)}{b+a\cos\theta}d\theta\\
&=\int_{\frac{\pi}2}^{\pi}\left(-a\cos\theta+b+\frac{a^2-b^2}{b+a\cos\theta}\right)d\theta\end{align}$$
Now that last term may look a little intimidating, but I just did it this morning, so we get$$\begin{align}\int_0^a\frac{\sqrt{a^2-x^2}}{b-x}dx&=\left[-a\sin\theta+b\theta+\frac{a^2-b^2}{\sqrt{b^2-a^2}}\cos^{-1}\left(\frac{b\cos\theta+a}{b+a\cos\theta}\right)\right]_{\frac{\pi}2}^{\pi}\\
&=a+b\frac{\pi}2+\sqrt{b^2-a^2}\left(\cos^{-1}\left(\frac ab\right)-\pi\right)\end{align}$$
</s>