Integral taken from instagram user @integralsforyou

$$\int\tan^2(x)\cdot\sin(x)dx$$

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

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

let $u=\cos(x)$

$du=-\sin(x)dx$

$$=-\int\frac{1-u^2}{u^2}du$$

$$=-\int(\frac{1}{u^2}-\frac{u^2}{u^2})du$$

$$=\int(1-\frac{1}{u^2})du$$

$$=u+\frac{1}{u}+C$$

$$=\cos(x)+\sec(x)+C$$