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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.

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  • $\begingroup$ Can we assume $|b|\gt |a|$? $\endgroup$ – abiessu Apr 14 '16 at 16:01
  • $\begingroup$ With the substitution $x=a \sin t$ the integral transforms to $\int_0^{\pi/2}\left[\frac{a^2-b^2}{b-a\sin t}+b+a\sin t\right]dt$ $\endgroup$ – vnd Apr 14 '16 at 16:34
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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$$\begin{align}\int_0^a\frac{\sqrt{a^2-x^2}}{b-x}dx&=\left[-a\sin\theta+b\theta+\frac{a^2-b^2}{\sqrt{b^2-a^2}}\cos^{-1}\left(\frac{b\cos\theta+a}{b+a\cos\theta}\right)\right]_{\frac{\pi}2}^{\pi}\\ &=a+b\frac{\pi}2+\sqrt{b^2-a^2}\left(\cos^{-1}\left(\frac ab\right)-\pi\right)\end{align}$$

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