Question

$\int \tan ({\sin }^{-1}x)dx$ is equal to

• A. $\frac{1}{\sqrt{1-x^2}}+c$
• B. $\sqrt{1-x^2}+c$
• C. $\frac{-x}{\sqrt{1-x^2}}+c$
• D. $\frac{x}{\sqrt{1-x^2}}+c$
• E. $-\sqrt{1-x^2}+c$

Answer 1

Answer: (E)

Let $I=\int \tan ({\sin }^{-1}x)dx$
$=\int \tan \left({\tan }^{-1}\frac{x}{\sqrt{1-x^2}}\right)dx$
$=\int \frac{x}{\sqrt{1-x^2}}dx$ Put $1-x^2=t^2\Rightarrow -2xdx=2tdt$
$\therefore$ $I=-\int \frac{tdt}{t}=-t+c$
$\Rightarrow$ $I=-\sqrt{1-x^2}+c$