Question

If $I_n=\int {\cos }^{n}xdx$, then show that $I_n=\frac{1}{n}{\cos }^{n-1}x\sin x+\frac{n-1}{n}{I}_{n-2}$ and from $n\ge 2$ deduce the value of $\int {\cos }^{4}xdx$.

Answer 1

Given:- $I_n=\int {\cos }^{n}xdx$

To prove:- $I_n=\frac{1}{n}\sin x{\cos }^{n-1}x-\frac{n-1}{n}{I}_{n-2}$

Proof:-

$I_n=\int {\cos }^{n}xdx$

$\Rightarrow I_n=\int \cos x\left({\cos }^{n}x\right)dx$

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

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

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

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

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

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

$\Rightarrow I_n=\frac{1}{n}\sin x{\cos }^{n-1}x-\frac{n-1}{n}{I}_{n-2}$

Hence proved.

Now,

$\int {\cos }^{4}xdx=I_4$

$I_4=\frac{1}{4}\sin x{\cos }^{3}x-\frac{3}{4}{I}_2.....\left(1\right)$

Now,

$I_2=\frac{1}{2}\sin x\cos x-\frac{1}{2}{I}_0$

$\Rightarrow I_2=\frac{1}{2}(\sin x\cos x-x+C)(\because I_0=\int dx=x+C)$

Substituting the value of $I_2$ in equation $\left(1\right)$, we have

$I_4=\frac{1}{4}\sin x{\cos }^{3}x-\frac{3}{4}\left(\frac{1}{2}(\sin x\cos x-x+C)\right)$

$I_4=\frac{1}{4}\sin x{\cos }^{3}x-\frac{3}{2}\sin x\cos x-\frac{3}{8}x+C$