Question

evaluate the integral.
int 7sin^{2}xcos^{2}x dx
int 7sin^{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 the integral $\int7\sin^{2}x\cos^{2}x dx=\int7\times\frac{1}{4}\times\frac{1 - \cos(4x)}{2}dx$.
$$\int7\sin^{2}x\cos^{2}x dx=\frac{7}{8}\int(1-\cos(4x))dx$$

Step2: Integrate term - by - term

The integral of a difference is the difference of the integrals. $\int(1-\cos(4x))dx=\int 1dx-\int\cos(4x)dx$. The integral of $1$ with respect to $x$ is $x$, and for $\int\cos(4x)dx$, let $u = 4x$, then $du=4dx$ and $\int\cos(4x)dx=\frac{1}{4}\sin(4x)$. So $\frac{7}{8}\int(1 - \cos(4x))dx=\frac{7}{8}(x-\frac{1}{4}\sin(4x))+C$.
$$\frac{7}{8}(x-\frac{1}{4}\sin(4x))+C=\frac{7}{8}x-\frac{7}{32}\sin(4x)+C$$

Answer:

$\frac{7}{8}x-\frac{7}{32}\sin(4x)+C$