Question

$\int \frac{dx}{({\sin }^{-1}x{)}^3\sqrt{1-x^2}}=$

• A. $\frac{-1}{2({\sin }^{-1}x{)}^2}+c$
• B. $\frac{1}{2({\sin }^{-1}x{)}^2}+c$
• C. $\frac{1}{3({\sin }^{-1}x{)}^3}+c$
• D. $\frac{1}{4({\sin }^{-1}x{)}^2}+c$

Answer 1

Answer: (A)

$\frac{-1}{2({\sin }^{-1}x{)}^2}+c$

$I=\int \frac{dx}{({\sin }^{-1}x{)}^3\sqrt{1-x^2}}$
Put ${\sin }^{-1}x=t$
$\Rightarrow \frac{1}{\sqrt{1-x^2}}dx=dt$
So, $I=\int \frac{dt}{t^3}$
$=\frac{t^{-2}}{-2}+C$
$I=-\frac{1}{2{({\sin }^{-1}x)}^2}+C$