Question

$\int \sin \left({\tan }^{-1}\sqrt{x}\right)dx(x\ge 0)$ is

• A. $\sqrt{x^2+x}-\frac{1}{2}\ln \left\{\right(x+\frac{1}{2})+\sqrt{x^2+x}\}+c$
• B. $2\sqrt{x^2+x}-\frac{1}{2}\ln \left\{\right(x+\frac{1}{2})+\sqrt{x^2+x}\}+c$
• C. $\sqrt{x^2+x}-\ln \left\{\right(x+\frac{1}{2})+\sqrt{x^2+x}\}+c$
• D. $\sqrt{x^2+x}+\ln \left\{\right(x+\frac{1}{2})+\sqrt{x^2+x}\}+c$

Answer 1

The correct option is A $\sqrt{x^2+x}-\frac{1}{2}\ln \left\{\right(x+\frac{1}{2})+\sqrt{x^2+x}\}+c$

$\int \sin \left({\tan }^{-1}\sqrt{x}\right)dx$
${\tan }^{-1}\sqrt{x}=\theta \Rightarrow \tan \theta =\sqrt{x}\int \frac{\sqrt{x}}{\sqrt{1+x}}dx=\int \frac{x}{\sqrt{x+x^2}}dx$
$=\frac{1}{2}\int \frac{(2x+1)-1}{\sqrt{x+x^2}}dx$
$=\frac{1}{2}[2\sqrt{x+x^2}-\ln \{(x+\frac{1}{2})+\sqrt{x^2+x}\left\}\right]+c$