Question

Prove that $\int {\sin }^{17}xdx=-\frac{{\sin }^{16}x\cos x}{17}+\frac{16}{17}\int {\sin }^{15}xdx+c$.

Answer 1

$\int {\sin }^{17}xdx$

Integration by parts when ${\sin }^{16}x$ is the differentiating term and $\sin x$ is integrating term
$={\sin }^{16}x\int \sin xdx-\int \frac{d}{dx}\left({\sin }^{16}x\right)(\int \sin xdx)dx={\sin }^{16}x(-\cos x)+\int 16{\sin }^{15}x{\cos }^{2}xdx=-{\sin }^{16}x\cos x+16\int {\sin }^{15}x(1-{\sin }^{2}x)dx=-{\sin }^{16}x\cos x+16\int {\sin }^{15}x-16\int {\sin }^{17}xdx17\int {\sin }^{17}xdx=-{\sin }^{16}x\cos x+16\int {\sin }^{15}x\int {\sin }^{17}xdx=-\frac{{\sin }^{16}x\cos x}{17}+\frac{16}{7}\int {\sin }^{15}x+C$

Hence proved