Question

Find $\int \frac{{\sin }^{6}x}{{\cos }^{8}x}dx$

Answer 1

$I=\int \frac{{\sin }^{6}x}{{\cos }^{8}x}dx$

$I=\int {\tan }^{6}x{\sec }^{2}xdx$

Put $\tan x=t$

${\sec }^{2}xdx=dt$

Therefore,

$I=\int t^{6}dt=\frac{t^7}{7}+C=\frac{{\tan }^{7}x}{7}+C$