Как решить интеграл cos^4x/sinx?

Решите задачу:

$\int \frac{cos^4 x}{sin x} dx=\\\\-\int \frac{cos^4 x (-sin x) dx }{sin^ 2x}=\\\\-\int \frac{cos^4 x d (cos x)}{1-cos^ 2x}=|cos x=t|=\\\\\int \frac{-t^4 dt}{1-t^2}=\\\\\int \frac{(1-t^4-1)dt}{1-t^2}=\\\\\int \frac{(1-t^2)(1+t^2) dt}{1-t^2}-\frac{dt}{1-t^2}=\\\\\int(1+t^2)dt-\int\frac{dt}{1-t^2}=t+\frac{t^3}{3}-\frac{1}{2}ln|\frac{1+t}{1-t}}|+C=\\\\cos x+\frac{cos^3 x}{3}-\frac{1}{2}ln|\frac{1+cos x}{1-cos x}|+C$