Question

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

Answer 1

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

$=\int \frac{{\left({\sin }^{2}x\right)}^3+{\left({\cos }^2\right)}^3}{{\sin }^{2}x{\cos }^{2}x}dx$

$\because {(a+b)}^3=a^3+b^3+3ab(a+b)$

$a^3+b^3={(a+b)}^3-3ab(a+b)$

$\because {\left({\sin }^{2}x\right)}^3+{\left({\cos }^2\right)}^3={({\sin }^{2}x+{\cos }^{2}x)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)$

$=1^3-3\sin x\cos x\left(1\right)$

$=1-3\sin x\cos x$

Therefore,

$\int \frac{{\left({\sin }^{2}x\right)}^3+{\left({\cos }^2\right)}^3}{{\sin }^{2}x{\cos }^{2}x}dx=\int \left(\frac{1-3{\sin }^{2}x{\cos }^{2}x}{si{n}^{2}x{\cos }^{2}x}\right)dx$

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

$=\int (\frac{({\sin }^{2}x+{\cos }^{2}x)}{si{n}^{2}x{\cos }^{2}x}-3)dx$

$=\int (\frac{{\sin }^{2}x}{si{n}^{2}x{\cos }^{2}x}+\frac{{\cos }^{2}x}{si{n}^{2}x{\cos }^{2}x}-3)dx$

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

$=\int ({\sec }^{2}x+\cos e{c}^{2}x-3)dx$

$=\int {\sec }^{2}xdx+\int \cos e{c}^{2}xdx-\int 3dx$

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

Hence, the value of integral is $\tan x-\cot x-3x+c$.