Question

Obtain the reduction formula for $I_n=\int si{n}^{n}xdx$ for an integer $n\ge 2$ and hence find $\int si{n}^{5}xdx$.

Answer 1

$\int {\sin }^{n}x$

$=\int {\sin }^{n-1}x\sin x$

$=\int {\sin }^{n-1}xd(-\cos x)$

$={\sin }^{n-1}x\times (-\cos x)-\int -\cos x(n-1){\sin }^{n-2}x\cos xdx$

$=-{\sin }^{n-1}x\cos x+\int {\cos }^{2}x(n-1){\sin }^{n-2}xdx$

$=-{\sin }^{n-1}x\cos x+(n-1)int{\sin }^{n-1}x(1-{\sin }^{2}x)dx$

$=-{\sin }^{n-1}x\cos x+(n-1)[\int {\sin }^{n-2}xdx-\int {\sin }^{n}xdx]$

So $I_n=-{\sin }^{n-1}x\cos x+(n-1)[I_{n-2}-I_n]$

$(n-1)I_n+I_n=-{\sin }^{n-1}x\cos x+(n-1)I_{n-2}$

$n{I}_n=-{\sin }^{n-1}x\cos x+(n-1)I_{n-2}$

$I_n=\frac{-{\sin }^{n-1}x\cos x}{n}+\frac{(n-1)I_{n-2}}{n}$

$I_5=\frac{-{\sin }^{4}x\cos x}{5}+\frac{4I_3}{5}$

$I_5=\frac{-{\sin }^{4}x\cos x}{5}+\frac{4}{5}(\frac{-{\sin }^{2}x\cos x}{3}+\frac{2}{3}{I}_1)$

$I_5=\frac{-{\sin }^{4}x\cos x}{5}+\frac{4}{5}(\frac{-{\sin }^{2}x\cos x}{3}-\frac{2}{3}\cos x)+c$