evaluate the following integral.\n\\(\\int7\\sin^{2}8x\\mathrm{d}x\\)\n\\(\\int7\\sin^{2}8x\\mathrm{d}x =…

Explanation:

Step1: Use double - angle formula

Recall $\sin^{2}\theta=\frac{1 - \cos2\theta}{2}$. Here $\theta = 8x$, so $7\sin^{2}8x=7\times\frac{1-\cos(16x)}{2}=\frac{7}{2}-\frac{7\cos(16x)}{2}$.

Step2: Integrate term - by - term

$\int\left(\frac{7}{2}-\frac{7\cos(16x)}{2}\right)dx=\int\frac{7}{2}dx-\frac{7}{2}\int\cos(16x)dx$. The integral of a constant $\int\frac{7}{2}dx=\frac{7}{2}x + C_1$. For $\int\cos(16x)dx$, let $u = 16x$, then $du=16dx$ and $\int\cos(16x)dx=\frac{1}{16}\int\cos(u)du=\frac{1}{16}\sin(u)+C_2=\frac{1}{16}\sin(16x)+C_2$. So $-\frac{7}{2}\int\cos(16x)dx=-\frac{7}{32}\sin(16x)+C_3$.

Step3: Combine results

$\int7\sin^{2}8x dx=\frac{7}{2}x-\frac{7}{32}\sin(16x)+C$, where $C = C_1 + C_3$.

Answer:

$\frac{7}{2}x-\frac{7}{32}\sin(16x)+C$