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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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lvk
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2 Answers2

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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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user5713492
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2

Note that $$\int_0^a {\frac{{\sqrt {{a^2} - {x^2}} }}{{b - x}}}dx= \int_0^a\frac{b+x}{\sqrt{a^2-x^2}}-\frac{b^2-a^2}{(b-x)\sqrt{a^2-x^2}}\ dx $$

where $\int_0^a\frac{b+x}{\sqrt{a^2-x^2}}dx = a+\frac{b\pi}2$ \begin{align} \int_0^a\frac{\sqrt{b^2-a^2}}{(b-x)\sqrt{a^2-x^2}}dx = &\ \tan^{-1}\frac{a^2-bx}{\sqrt{(b^2-a^2)(a^2-x^2)}}\bigg|_0^a =\tan^{-1}\frac{\sqrt{b^2-a^2}}a \end{align}

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