Answer
$$\int\frac{\tan^2x}{\csc x}dx=\cos x+\frac{1}{\cos x}+C$$
Work Step by Step
$$A=\int\frac{\tan^2x}{\csc x}dx=\int\frac{\frac{\sin^2x}{\cos^2x}}{\frac{1}{\sin x}}dx=\int\frac{\sin^3x}{\cos^2x}dx$$ $$A=\int\frac{\sin^2x}{\cos^2x}(\sin xdx)$$ $$A=-\int\frac{1-\cos^2x}{\cos^2x}d(\cos x)$$
We set $u=\cos x$
$$A=-\int\frac{1-u^2}{u^2}du=\int\frac{u^2-1}{u^2}du$$ $$A=\int du-\int\frac{1}{u^2}du$$ $$A=u+\frac{1}{u}+C$$ $$A=\cos x+\frac{1}{\cos x}+C$$
