^ -1 x a+x dx Hint: Put x = a ^2θ - Vidyadip

Question

$\int\sin^{-1}\sqrt{\frac{\text{x}}{\text{a}+\text{x}}}\text{dx}$
Hint: Put $\text{x} = \text{a}\tan^2θ$

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Answer

Let $\text{I}=\int\sin^{-1}\sqrt{\frac{\text{x}}{\text{a}+\text{x}}}\text{dx}$
Put $\text{x}=\text{a}\tan^2\theta$
$\Rightarrow\ \text{dx}=2\text{a}\tan\theta\sec^2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\sin^{-1}\sqrt{\frac{\text{a}\tan^2\theta}{\text{a}+\text{a}\tan^2\theta}}(2\text{a}\tan\theta\cdot\sec^2\theta)\text{d}\theta$
$=2\text{a}\int\sin^{-1}\Big(\frac{\tan\theta}{\sec\theta}\Big)\tan\theta\cdot\sec^2\theta\text{ d}\theta$
$=2\text{a}\int\sin^{-1}(\sin\theta)\tan\theta\cdot\sec^2\theta\text{ d}\theta$
$=2\text{a}\int\theta\cdot\tan\theta\sec^2\theta\text{ d}\theta\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{I} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ II}$
$=2\text{a}\bigg[\theta\cdot\int\tan\theta\cdot\sec^2\theta\text{ d}\theta-\int\Big(\frac{\text{d}}{\text{d}\theta}\theta\cdot\int\tan\theta\cdot\sec^2\theta\text{ d}\theta\Big)\text{d}\theta\bigg]$
$\begin{bmatrix} \text{let} \tan\theta=\text{t} \\\sec^2\theta\text{d}\theta=\text{dt}\\ \ \ \ \ \ \int\tan\theta\sec^2\theta\text{ d}\theta=\int\text{tdt}=\frac{\text{t}^2}{2}=\frac{\tan^2\theta}{2}\end{bmatrix}$
$=2\text{a}\Big[\theta\cdot\frac{\tan^2\theta}{2}-\int\frac{\tan^2\theta}{2}\text{d}\theta\Big]$
$=\text{a}\theta\tan^2\theta-\text{a}\int(\sec^2\theta-1)\text{d}\theta$
$=\text{a}\theta\cdot\tan^2\theta-\text{a}\tan\theta+\text{a}\theta+\text{C}$
$=\text{a}\bigg[\frac{\text{x}}{\text{a}}\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}+\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}\bigg]+\text{C}$

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