1. $int\frac{sin^{3}x}{cos^{2}x}dx$

Explanation:

Step1: Use substitution

Let $u = \cos x$, then $du=-\sin xdx$.

Step2: Rewrite the integral

The integral $\int\frac{\sin x}{\cos^{3}x}dx=-\int\frac{du}{u^{3}}$.

Step3: Integrate the new - form

We know that $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $n=-3$, $-\int u^{-3}du=-\frac{u^{-3 + 1}}{-3 + 1}+C$.

Step4: Simplify the result

$-\frac{u^{-2}}{-2}+C=\frac{1}{2u^{2}}+C$.

Step5: Substitute back $u = \cos x$

The result is $\frac{1}{2\cos^{2}x}+C$.

Answer:

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