Question

If $\int {\sin }^{2}x{\cos }^{2}xdx=\frac{x}{A}+\frac{1}{B}\sin 4x+C$, then the value of $A-B$ is equal to
(where $A,B$ are fixed constants and $C$ is integration constant)

Answer 1

$I=\int {\sin }^{2}x{\cos }^{2}xdx\Rightarrow I=\frac{1}{4}\int {\left(\sin 2x\right)}^{2}dx\Rightarrow I=\frac{1}{4}\int \frac{1-\cos 4x}{2}dx[\because 1-\cos 2\theta =2{\sin }^2\theta ]\Rightarrow I=\frac{1}{8}\int dx-\frac{1}{8}\int \cos 4xdx\Rightarrow I=\frac{x}{8}-\frac{1}{32}\sin 4x+C$
So,
$A=8,B=-32\therefore A-B=40$