Question

Solve: $\int {\cos }^{5}x.cose{c}^x.dx$

Answer 1

$\int {\cos }^{5}x.cosecxdx=\int \frac{{cos}^{5}x}{sinx}dx(cosecx=\frac{1}{sinx})=\int \frac{{cos}^{4}x}{sinx}cosxdx=\int \frac{{\left({cos}^{2}x\right)}^2}{sinx}cosxdx=\int \frac{{(1-{sin}^{2}x)}^2}{sinx}cosxdx({cos}^{2}x=1-{sin}^{2}x)=\int (\frac{1}{sinx}-2sinx+{sin}^{3}x)cosxdxLetsinx=t\Rightarrow cosxdx=dt\therefore \int {cos}^{5}x.cosecxdx=\int (\frac{1}{t}-2t+t^3)dt=logt-t^2+\frac{t^3}{3}=log\left(sinx\right)-{sin}^{2}x+\frac{{sin}^{3}x}{3}+C$