evaluate the following integral.\n int sin^{2}xcos^{2}x dx \n int sin^{2}xcos^{2}x dx = square

Explanation:

Step1: Use double - angle formulas

Recall that $\sin(2x) = 2\sin x\cos x$, so $\sin x\cos x=\frac{1}{2}\sin(2x)$ and $\sin^{2}x\cos^{2}x=\frac{1}{4}\sin^{2}(2x)$. Also, $\sin^{2}\alpha=\frac{1 - \cos(2\alpha)}{2}$, so $\sin^{2}(2x)=\frac{1-\cos(4x)}{2}$. Then $\sin^{2}x\cos^{2}x=\frac{1}{4}\times\frac{1 - \cos(4x)}{2}=\frac{1}{8}(1 - \cos(4x))$.

Step2: Integrate term - by - term

We know that $\int kdx=kx + C$ and $\int\cos(ax)dx=\frac{1}{a}\sin(ax)+C$ ($a\neq0$). [ \begin{align*} \int\sin^{2}x\cos^{2}xdx&=\int\frac{1}{8}(1 - \cos(4x))dx\ &=\frac{1}{8}\int(1 - \cos(4x))dx\ &=\frac{1}{8}\left(\int 1dx-\int\cos(4x)dx\right)\ &=\frac{1}{8}\left(x-\frac{1}{4}\sin(4x)\right)+C \end{align*} ]

Answer:

$\frac{1}{8}\left(x-\frac{1}{4}\sin(4x)\right)+C$