Question

Integrate the following functions with respect to x:
$\frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}$

Answer 1

$\frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \frac{({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x{\cos }^{2}x}$
$(\because a^3+b^3=(a+b{)}^3-3ab(a+b))$
$=\int \frac{1-3{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int (\frac{1}{{\sin }^{2}x{\cos }^{2}x}-3)dx$
$=\int (\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}-3)dx$
$=\int {\sec }^{2}xdx+\int cose{c}^{2}xdu-3\int dx$
$=\tan x-\cot x-3x+C$$\frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \frac{({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x{\cos }^{2}x}$
$(\because a^3+b^3=(a+b{)}^3-3ab(a+b))$
$=\int \frac{1-3{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int (\frac{1}{{\sin }^{2}x{\cos }^{2}x}-3)dx$
$=\int (\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}-3)dx$
$=\int {\sec }^{2}xdx+\int cose{c}^{2}xdu-3\int dx$
$=\tan x-\cot x-3x+C$