Question

Find the integrals of the functions.
$\int \frac{1}{sinxco{s}^{3}x}dx.$

Answer 1

Let $I=\int \frac{1}{sinxco{s}^{3}x}dx=\int \frac{cosx}{sinx}se{c}^{4}xdx$
$=\int \frac{se{c}^{2}xse{c}^{2}x}{tanx}dx=\int \frac{(1+ta{n}^{2}x)se{c}^{2}x}{tanx}dx(\because se{c}^{2}x=1+ta{n}^{2}x)\text{Putting}tanx=t\Rightarrow se{c}^{2}x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{se{c}^{2}x}\therefore I=\int \left[\frac{1+t^2}{t}\right]se{c}^{2}x\frac{dt}{se{c}^{2}x}=\int \frac{1+t^2}{t}dt=\int [\frac{1}{t}+t]dt=log|t|+\frac{t^2}{2}+C\Rightarrow log|tanx|+\frac{ta{n}^{2}x}{2}+C$