Question

Solve:$\int ({\sin }^{2}x{\cos }^2{x}_{)}dx$

Answer 1

We know that,

$\sin 2\theta =2\sin \theta \cos \theta \Rightarrow \sin \theta \cos \theta =\frac{1}{2}\left(\sin 2\theta \right)$

and $\sin 2\theta =2\sin \theta \cos \theta \Rightarrow \sin \theta \cos \theta =\frac{1}{2}\left(\sin 2\theta \right)$

$\therefore \int {\sin }^{2}x{\cos }^{2}xdx=\int {\left(\sin x\cos x\right)}^{2}dx=\int {\left(\frac{1}{2}\sin 2x\right)}^{2}dx=\frac{1}{4}\int \left({\sin }^{2}2x\right)dx=\frac{1}{4}\int \frac{1}{2}[1-\cos (2\cdot 2x)]dx=\frac{1}{8}\int (1-\cos 4x)dx=\frac{1}{8}[\int \left(1\right)dx-\int \cos 4xdx]=\frac{1}{8}(x-\frac{1}{4}\sin 4x)+C=\frac{x}{8}-\frac{\sin 4x}{32}+C\therefore \int {\sin }^{2}x{\cos }^{2}xdx=\frac{x}{8}-\frac{\sin 4x}{32}+C.$