The value of $\int \sqrt[3]{\frac{\sin ^n x}{\cos ^{n+6} x}} d x$, is
Answer & explanation
Correct answer: option 2
We have,
$I=\int \sqrt[3]{\frac{\sin ^n x}{\cos ^{n+6} x}} d x=\int 3 \sqrt{\tan ^n x} \sec ^2 x d x$
$\Rightarrow I=\int \tan ^{n / 3} x d(\tan x)=\frac{(\tan x)^{\frac{n}{3}+1}}{\frac{n}{3}+1}=\frac{3}{n+3} \tan ^{\frac{n}{3}+1} x+C$