Question
Evaluate the following:
$\int\sqrt{\frac{\text{a}+\text{x}}{\text{a}-\text{x}}}\text{dx}$

Answer

Let $\text{I}=\int\sqrt{\frac{\text{a}+\text{x}}{\text{a}-\text{x}}}\text{dx}$
Put $\text{x}=\text{a}\cos2\theta$ $\Rightarrow\ =\text{a}\cdot\sin2\theta\cdot2\cdot\text{d}\theta$
$\therefore\ =-2\int\sqrt{\frac{\text{a}+\text{a}\cos2\theta}{\text{a}-\text{a}\cos2\theta}}\cdot\text{a}\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\sqrt{\frac{1+\cos2\theta}{1-\cos2\theta}}\sin2\theta\text{ d}\theta$
$=-2\int\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\cot\theta\cdot\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\frac{\cos\theta}{\sin\theta}\cdot2\sin\theta\cos\theta\text{ d}\theta$
$=-4\text{a}\int\cos^2\theta\text{ d}\theta$
$=-2\text{a}\int(1+\cos2\theta)\text{d}\theta$
$=-2\text{a}\Big[\theta+\frac{1}{2}\sin2\theta\Big]+\text{C}$
$=-2\text{a}\bigg[\frac{1}{2}\cos^{-1}\frac{\text{x}}{\text{a}}+\frac{1}{2}\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\bigg]+\text{C}$
$=-\text{a}\bigg[\cos^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\bigg]+\text{C}$

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