Answer
$\int$$\frac{cos^4 x}{sin^6x}dx=-\frac{(\cot x)^5}{5}+C$
Work Step by Step
$\int$$\frac{cos^4 x}{sin^6x}dx$
First, split the integral.
$\int$$\frac{cos^4 x}{sin^4x}\times\frac{1}{sin^2x}dx$
$\int$$cot ^4x \times csc^2x dx$
Then, Let $u= cot x$ and $du= -csc^2 x$ .
$\int-u^4 du$
$-\frac{u^5}{5}$
Subtitute back $u=\cot x $ to form a final answer.
$-\frac{(\cot x)^5}{5}+C$
