Question

Solve:$\int {\cos }^{n}x.\sin nx.dx$

Answer 1

$I_{m,n}=\int {cos}^{m}x.\sin nxdx={\cos }^{m}x.\int \sin nxdx-\int [\frac{d}{dx}{cos}^{m}x.\int \sin nxdx]dx=-\left(\frac{{\cos }^{m}x.\cos nx}{n}\right)-\int \frac{m}{n}{\cos }^{m-1}x.\sin x.\left(\frac{-\cos nx}{n}\right)dx=-\left(\frac{{\cos }^{m}x.\cos nx}{n}\right)+\frac{m}{n}\int {\cos }^{m-1}x.(\sin x.\cos nx)dx⟶\left(1\right)\because \sin (n-1)x=\sin nx.\cos x-\sin x\cos nx\Rightarrow \sin x\cos nx=\sin nx.\cos x-\sin (n-1)x⟶\left(2\right)puttingthevalueof\left(2\right)inequation\left(1\right)I_{m,n}=-\left(\frac{{\cos }^{m}x.\cos nx}{n}\right)+\frac{m}{n}\int {\cos }^{m-1}x.(\sin nx.\cos x-\sin (n-1)x)dx=-\left(\frac{{\cos }^{m}x.\cos nx}{n}\right)+\frac{m}{n}\int {\cos }^{m}x.\sin nxdx-\frac{m}{n}\int {\cos }^{m-1}x.\sin (n-1)xdx=-\left(\frac{{\cos }^{m}x.\cos nx}{n}\right)+\frac{m}{n}{I}_{m,n}-\frac{m}{n}{I}_{m-1,n-1}\Rightarrow \frac{m-n}{n}{I}_{m,n}=\left(\frac{{\cos }^{m}x.\cos nx}{n}\right)+\frac{m}{n}{I}_{m-1,n-1}\Rightarrow I_{m,n}=\frac{{\cos }^{m}x.\cos nx}{m-n}+\frac{m}{m-n}{I}_{m-1,n-1}$