Question

∫sin−1√xa+xdx

• A. $\left(a+x\right)arc\tan \sqrt{\frac{x}{a}}-\sqrt{ax}+c$
• B. $\left(a+x\right)arc\tan \sqrt{\frac{x}{a}}+\sqrt{ax}+c$
• C. $\left(a-x\right)arc\tan \sqrt{\frac{x}{a}}-\sqrt{ax}+c$
• D. $\left(a+x\right)arc\cot \sqrt{\frac{x}{a}}-\sqrt{ax}+c$

Answer 1

The correct option is A $\left(a+x\right)arc\tan \sqrt{\frac{x}{a}}-\sqrt{ax}+c$

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

Put $x=a{\tan }^2\theta$

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

$\therefore I=\int {\sin }^{-1}\sqrt{{\sin }^2\theta }2a{\sec }^2\theta \tan \theta d\theta$

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

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

Let $u=\theta \Rightarrow du=d\theta$

$dv={\sec }^2\theta \tan \theta d\theta$

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

Let $t=\tan \theta \Rightarrow dt={\sec }^2\theta d\theta$

$v=\int tdt=\frac{t^2}{2}=\frac{{\tan }^2\theta }{2}$

$\therefore I=2a\left[\theta \int {\sec }^2\theta \tan \theta d\theta -\int {\left(\theta \right)}^{\prime }\left(\int {\sec }^2\theta \tan \theta d\theta \right)d\theta \right]+C$

$=2a[\theta \frac{{\tan }^2\theta }{2}-\int \frac{{\tan }^2\theta }{2}d\theta ]+C$

$=a[\theta {\tan }^2\theta -\int ({\sec }^2\theta -1)d\theta ]+C$

$=a[a{\tan }^2\theta -\tan \theta +\theta ]+C$

$=a[\frac{x}{a}{\tan }^{-1}\sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}+{\tan }^{-1}\sqrt{\frac{x}{a}}]+C$

$=a[(\frac{x}{a}+1){\tan }^{-1}\sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}]+C$

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

$=(x+a){\tan }^{-1}\sqrt{\frac{x}{a}}-a\sqrt{\frac{x}{a}}\times \sqrt{\frac{a}{a}}+C$

$=(a+x){\tan }^{-1}\sqrt{\frac{x}{a}}-\sqrt{ax}+C$