Question

If the value of integral $\int {\sin }^{4}x{\cos }^{2}xdx=\frac{x}{p}+\frac{\sin 2x}{q}+\frac{\sin 4x}{r}+\frac{\sin 6x}{s}+C,$ for fixed constants $p,q,r$ and $s.$ Then the value of $\frac{p+q+r+s}{4}=$
(where $C$ is integration constant)

Answer 1

Let
$I=\int {\sin }^{4}x{\cos }^{2}xdx=\frac{1}{4}\int {\sin }^{2}x{\sin }^{2}2xdx=\frac{1}{16}\int (1-\cos 2x)(1-\cos 4x)dx(\because 2{\sin }^{2}x=1-\cos 2x)=\frac{1}{16}\int (1-\cos 2x-\cos 4x+\cos 2x\cos 4x)dx=\frac{1}{16}\int (1-\cos 2x-\cos 4x+\frac{\cos 6x}{2}+\frac{\cos 2x}{2})(\because 2\cos a\cos b=\cos (a+b)+\cos (a-b\left)\right)=\frac{1}{16}[x-\frac{\sin 2x}{4}-\frac{\sin 4x}{4}+\frac{\sin 6x}{12}]+C$
$\Rightarrow \frac{p+q+r+s}{4}=\frac{16(1-4-4+12)}{4}=20$