Question

Integrate:
$\int \frac{\cos 4x}{{\sin }^{2}x}dx$

Answer 1

$I=\int \frac{\cos 4x}{{\sin }^{2}x}dx$

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

$=\int \frac{({\cos }^{2}x-{\sin }^{2}x{)}^2-4{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x}dx$

$=\int \frac{{\cos }^{4}x+{\sin }^{4}x-6{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x}dx$

$=\int \frac{{\sin }^{4}x+{\cos }^{2}x(1-7{\sin }^{2}x)dx}{{\sin }^{2}x}$

$[\because {\sin }^{2}x+{\cos }^{2}x=1]$

$\int {\sin }^{2}xdx+\int \frac{{\cos }^{2}x(1-7{\sin }^{2}x)dx}{{\sin }^{2}x}$

$=I_1+I_2$ ............. $\left(1\right)$

$I_1=\int {\sin }^{2}xdx=\int \frac{(1-\cos 2x)}{2}dx$

$=\frac{x}{2}-\frac{\sin 2x}{4}$ ....... $\left(2\right)$

$I_2=\int \frac{{\cos }^{2}x(1-7{\sin }^{2}x)}{{\sin }^{2}x}dx$

$=\int {\cot }^{2}dxd-7\int {\cos }^{2}xdx$

$=\int ({\csc }^{2}x-1)dx-\frac{7}{2}\int (1+\cos 2x)dx$

$=-\cot x-x\frac{-7}{2}x-\frac{7}{4}\sin 2x$

$=-\cot x-\frac{7}{4}\sin 2x-\frac{9}{2}x$ ........ $\left(3\right)$

$I=I_1+I_2$

$\Rightarrow I=-\cot x-2\sin 2x-4x+C$