Question

Obtain reduction formula for $I_n=\int {\sin }^{n}xdx,n$ being a positive integer, $n\ge 2$ and hence deduce the value of $\int {\sin }^{4}xdx$

Answer 1

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

now do integration by parts , we get $I_n={\sin }^{n-1}x(-\cos x)+{\int }^{}(n-1){\cos }^{2}x{\sin }^{n-2}x=-{\sin }^{n-1}x\cos x+(n-1){\int }^{}{\sin }^{n-2}x(1-{\sin }^{2}x)$

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

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

Now put $n=4$ , we get $4\int {\sin }^{4}xdx=3\int {\sin }^{2}xdx-{\sin }^{3}x\cos x=3\left(\frac{x-\sin x\cos x}{2}\right)-{\sin }^{3}x\cos x$

Therefore $\int {\sin }^{4}xdx=\frac{1}{4}(3\left(\frac{x-\sin x\cos x}{2}\right)-{\sin }^{3}x\cos x)$