Question

Evaluate:
$\int \frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}dx$

Answer 1

$\int \frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}dx$

$=\int \frac{1-3{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$

$=\int cose{c}^{2}x{\sec }^{2}xdx-\int 3dx$

$=\int (1+{\cot }^{2}x)(1+{\tan }^{2}x)dx-3x+C$

$=x+\int {\tan }^{2}xdx+\int {\cot }^{2}xdx+x-3x+c$

$=-x+\int ({\sec }^{2}x-1)dx+\int (cose{c}^{2}x-1)dx$

$=\tan x-\cot x-3x+C$