Question

$I=\int {\cos }^{5}xdx$

Answer 1

$\int {\cos }^{5}xdx=\int {(1-{\sin }^{2}x)}^2\times \cos xdxlet\sin x=t\cos xdx=dt=\int {(1-t^2)}^{2}dt=\int (1+t^4-2t^2)dt=t+\frac{t^5}{5}-\frac{2t^3}{3}+C=\sin x+\frac{{\sin }^{5}x}{5}-\frac{2{\sin }^{3}x}{3}+C$

$(t=\sin x)$