evaluate the following integral.\n int 22sin^{2}xcos^{2}x dx \n int 22sin^{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, $\sin^{2}\alpha=\frac{1 - \cos2\alpha}{2}$, so $\sin^{2}2x=\frac{1-\cos4x}{2}$. Then $22\sin^{2}x\cos^{2}x = 22\times\frac{1}{4}\times\frac{1 - \cos4x}{2}=\frac{11}{4}(1-\cos4x)$.

Step2: Integrate term - by - term

We know that $\int kdx=kx + C$ and $\int\cos(ax)dx=\frac{1}{a}\sin(ax)+C$ ($a\neq0$, $k$ is a constant). $\int\frac{11}{4}(1 - \cos4x)dx=\frac{11}{4}\int(1 - \cos4x)dx=\frac{11}{4}\left(\int 1dx-\int\cos4xdx\right)$. $\int 1dx=x + C_1$ and $\int\cos4xdx=\frac{1}{4}\sin4x + C_2$. So $\frac{11}{4}\left(x-\frac{1}{4}\sin4x\right)+C=\frac{11}{4}x-\frac{11}{16}\sin4x + C$.

Answer:

$\frac{11}{4}x-\frac{11}{16}\sin4x + C$