Question:
The integral $\int \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$ is equal to:
Correct Option: 1
Solution:
$I=\int \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} d x$
$I=\int \frac{\sec ^{2} x d x}{\tan ^{\frac{4}{3}} x}$
Put $\tan x=z$
$\Rightarrow \sec ^{2} x d x=d z$
$\Rightarrow I=\int z^{-\frac{4}{3}} \cdot d z=\frac{z^{\frac{-1}{3}}}{\left(\frac{-1}{3}\right)}+C \Rightarrow I=-3(\tan x)^{\frac{-1}{3}}+C$