evaluate the following integral.\n int 3sin^{2}4x dx \n int 3sin^{2}4x dx=square

Explanation:

Step1: Use double - angle formula

Recall $\sin^{2}\theta=\frac{1 - \cos2\theta}{2}$. Here $\theta = 4x$, so $3\sin^{2}4x=3\times\frac{1-\cos(8x)}{2}=\frac{3}{2}-\frac{3}{2}\cos(8x)$.

Step2: Integrate term - by - term

$\int\left(\frac{3}{2}-\frac{3}{2}\cos(8x)\right)dx=\int\frac{3}{2}dx-\frac{3}{2}\int\cos(8x)dx$. The integral of a constant $\int\frac{3}{2}dx=\frac{3}{2}x$. For $\int\cos(8x)dx$, use substitution $u = 8x$, $du=8dx$, then $\int\cos(8x)dx=\frac{1}{8}\int\cos(u)du=\frac{1}{8}\sin(u)+C=\frac{1}{8}\sin(8x)+C$. So $-\frac{3}{2}\int\cos(8x)dx=-\frac{3}{16}\sin(8x)$.

Step3: Combine results

$\int3\sin^{2}4x dx=\frac{3}{2}x-\frac{3}{16}\sin(8x)+C$.

Answer:

$\frac{3}{2}x-\frac{3}{16}\sin(8x)+C$