Question

$I=\int {\cos }^{4}x.dx$

Answer 1

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

Now ${\cos }^{4}x={\cos }^{3}x.\cos x$

$\int {\cos }^{4}xdx=\int {\cos }^{3}xd\left(\sin x\right)$

$\int {\cos }^{4}xdx=\sin x{\cos }^{3}x-\int \sin xd\left({\cos }^{3}x\right)\to$ By parts

$\int {\cos }^{4}xdx=\sin x{\cos }^{3}x+3\int si{n}^{2}x{\cos }^{2}xdx$

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

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

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

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

Now, $\int {\cos }^{2}xdx=\int \left(\frac{\cos 2x+1}{2}\right)dx$

$\therefore \int {\cos }^{2}xdx=\frac{\sin 2x}{4}+\frac{x}{2}$

$\therefore \int {\cos }^{4}xdx=\frac{\sin x{\cos }^{3}x}{4}+\frac{3}{4}(\frac{\sin 2x}{4}+\frac{x}{2})$

$=\int {\cos }^{4}xdx=\frac{\sin x{\cos }^{3}x}{4}+\frac{3\sin 2x}{16}+\frac{3x}{8}$