Question

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

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

Answer 1

The correct option is B $\frac{1}{4}(si{n}^{-1}x{)}^4+c$

$\int \frac{(si{n}^{-1}x{)}^3}{\sqrt{1-x^2}}dx$
Let $si{n}^{-1}x=t$
Hence
$\frac{1}{\sqrt{1-x^2}}dx=dt$
Therefore the above integral transforms to
$\int t^{3}dt$
$=\frac{t^4}{4}+c$
$=\frac{(si{n}^{-1}x{)}^4}{4}+c$