Question

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

Answer 1

We have,

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

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

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

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

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

We know that

${\cos }^{2}x=\frac{1+\cos 2x}{2}$

Therefore,

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

$I=\tan x+\frac{1}{2}(x+\frac{\sin 2x}{2})-2x+C$

Hence, this is the answer.