The value of $\int \sqrt[3]{\frac{{\displaystyle {\sin }^{n}x}}{{\displaystyle {\cos }^{n+6}x}}}x,$ is 

Solution:

We have,

$I=\int \sqrt[3]{\frac{{\sin }^{n}x}{{\cos }^{n+6}x}}dx=\int 3\sqrt{{\tan }^{n}x}x{\sec }^{2}xdx$

$\Rightarrow l=\int {\tan }^{n/3}xd(\tan x)=\frac{(\tan x{)}^{\frac{n}{3}+1}}{{\displaystyle \frac{n}{3}+1}}=\frac{3}{n+3}{\tan }^{\frac{n}{3}+1}x+C$