Question

Obtain the reduction formula for $\int {\sin }^{n}xdx$ an integer $n\ge 2$ and deduce the value of $\int {\sin }^{4}xdx$.

Answer 1

Let $I_n=\int {\sin }^{n}xdx$

$\Rightarrow I_n=\int \sin x\cdot {\sin }^{n-1}xdx$

Integrating by parts, we have

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

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

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

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

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

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

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

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

Now,

$\int {\sin }^{4}xdx$

$=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{4}\int {\sin }^{2}xdx$

$=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{4}\int \frac{1-\cos 2x}{2}dx$

$=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{8}\int dx-\frac{3}{8}\int \cos 2xdx$

$=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{8}x-\frac{3}{8}\left(\frac{1}{2}\sin 2x\right)+C$

$=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{8}x-\frac{3}{16}\left(\sin 2x\right)+C$