# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$

Solution:

Let $I=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$

Let $\mathrm{x}=\mathrm{a} \tan ^{2} \theta$

$\mathrm{dx}=2 \mathrm{a} \tan ^{2} \theta \sec ^{2} \theta$

$I=\int\left(\sin ^{-1} \sqrt{\frac{\operatorname{atan}^{2} \theta}{a+\operatorname{atan}^{2} \theta}}\right) 2 a \tan ^{2} \theta \sec ^{2} \theta d \theta$

$=\int \sin ^{-1}(\sin \theta) 2 a \tan ^{2} \theta \sec ^{2} \theta d \theta$

$=\int 2 \theta a \tan ^{2} \theta \sec ^{2} \theta d \theta$

$=2 a \int \theta \tan ^{2} \theta \sec ^{2} \theta d \theta$

Using integration by parts,

$=2 \mathrm{a}\left(\theta \int \tan ^{2} \theta \sec ^{2} \theta \mathrm{d} \theta-\int 1 \int \tan ^{2} \theta \sec ^{2} \theta \mathrm{d} \theta\right)$

$=2 \mathrm{a}\left[\theta \frac{\tan ^{2} \theta}{2}-\int \frac{\tan ^{2} \theta}{2} \mathrm{~d} \theta\right]$

$=\mathrm{a} \theta \tan ^{2} \theta-\frac{2 \mathrm{a}}{2} \int\left(\sec ^{2} \theta-1\right) \mathrm{d} \theta$

$=a \theta \tan ^{2} \theta-a \tan \theta+a \theta+c$

$=a\left(\tan ^{-1} \sqrt{\frac{x}{a}}\right) \frac{x}{a}-a \sqrt{\frac{x}{a}}+a \tan ^{-1} \sqrt{\frac{x}{a}}+c$

$=x \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{a x}+a \tan ^{-1} \sqrt{\frac{x}{a}}+c$