Question

$\int \frac{3x^2}{\sqrt{(9-16x^6)}}dx.$

• A. $\frac{1}{4}{\sin }^{-1}\frac{4}{3}{x}^{3.}$
• B. $\frac{1}{2}{\sin }^{-1}\frac{4}{3}{x}^{3.}$
• C. $\frac{1}{4}{\sin }^{-1}\frac{2}{3}{x}^{3.}$
• D. $\frac{1}{4}{\sin }^{-1}\frac{4}{3}{x}^{6.}$

Answer 1

The correct option is A $\frac{1}{4}{\sin }^{-1}\frac{4}{3}{x}^{3.}$

Let $I=\int \frac{3x^2}{\sqrt{(9-16x^6)}}dx$
Put $4x^3=t\Rightarrow 12x^{2}dx=dt$
Therefore
$I=\frac{1}{4}\int \frac{dt}{\sqrt{(3^2-t^2)}}=\frac{1}{4}{\sin }^{-1}\frac{t}{3}=\frac{1}{4}{\sin }^{-1}\frac{4}{3}{x}^3$
Hence, option 'D' is correct.