# Evaluate the following:$\int\sin^{-1}\sqrt{\large\frac{x}{a+x}}dx\quad(Hint:Put\;x=a\tan^2\theta)$

Toolbox:
• If $f(x)$ is substituted by $f(t)$,then $f'(x)=f'(t).$Hence $\int f(x)dx=\int f(t)dt.$
• $\large\frac{d}{dx}\normalsize(\tan\theta)=\sec^2\theta.$
• $\int \sec^2\theta\; d\theta=\tan\theta+c$.
Step 1:
Let $I=\int\sin^{-1}\sqrt{\large\frac{x}{a+x}}\normalsize dx.$
Consider $\large\frac{x}{a+x}$
Put $x=a\tan^2\theta$.
On differentiating with respect to t we get,
$dx=2a\tan\theta.\sec^2\theta.dt$
$\sqrt{\large\frac{x}{a+x}}=\large \frac{a\tan^2\theta}{a+a\tan^2\theta}=\frac{a\tan^2\theta}{a(1+\tan^2\theta)}$
But $1+\tan^2\theta=\sec^2\theta$
Hence $\large\frac{x}{a+x}=\sqrt{\frac{a\tan^2\theta}{a\sec^2\theta}}=\sqrt{\frac{\sin^2\theta}{\cos^2\theta}\normalsize \times\cos^2\theta}=\normalsize\sin\theta$.
On substituting this we get,
$I=\int \sin^{-1}(\sin\theta)2a\tan\theta.\sec^2\theta.d\theta.$
$\;\;=2a\int\theta (\tan\theta.\sec^2\theta)d\theta.$
Hence $\sin^{-1}\sqrt{\large\frac{x}{a+x}}dx=2a\big(\theta.\large{\frac{\tan^2\theta}{2}}-\int\large{\frac{\tan^2\theta}{2}}.\normalsize d\theta\big).$
If $t=\tan\theta,$then $\int\tan\theta.\sec^2\theta.d\theta=\int tdt=\large\frac{t^2}{2}=\frac{\tan^2\theta}{2}$
Step 2:
$I=\large\frac{2a}{2}\normalsize [a\tan^2\theta-\int \tan^2\theta d\theta]$
But $\tan^2\theta=\sec^2\theta-1$
$I=a[\theta\tan^2\theta-\int(\sec^2\theta-1)d\theta]$
On integrating we get,
$I=a[\theta\tan^2\theta-(\tan\theta-\theta)+c]$
$\;\;=a[\theta(1+\tan^2\theta)-\tan\theta]+c.$
But $\tan^2\theta=\large\frac{x}{a}$ and $\theta=\tan^{-1}\sqrt{\large\frac{x}{a}}$
Hence $I=a[\tan^{-1}\sqrt{\large\frac{x}{a}}(1+\large\frac{x}{a})-\sqrt{\large\frac{x}{a}}]+c.$
On simplifying we get,
I=$(x+a)\tan^{-1}\sqrt{\large\frac{x}{a}}\normalsize -\sqrt{ax}+c$