Question

$\int \frac{1}{({\sin }^{-1}x{)}^2\sqrt{1-x^2}}dx$

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

Answer 1

The correct option is C $\frac{-1}{{\sin }^{-1}x}+c$

$\int \frac{1}{{\left({\sin }^{-1}x\right)}^2\sqrt{1-x^2}}.dx$

we know that

$d{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}.dx$

so, $\int {\frac{1}{{\sin }^{-1}x}}^2.d\left({\sin }^{-1}x\right)$

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

$=-\frac{1}{{\sin }^{-1}x}+c$

$\int \frac{1}{\left({\sin }^{-1}x\right)\sqrt{1-x^2}}dx$

$=-\frac{1}{{\sin }^{-1}x}+c$