evaluate $int 9x^{6}sin^{2}(x^{7})dx$ + c

Explanation:

Step1: Use substitution

Let $u = x^{7}$, then $du=7x^{6}dx$ and $x^{6}dx=\frac{1}{7}du$. The integral becomes $\int9\sin^{2}(u)\cdot\frac{1}{7}du=\frac{9}{7}\int\sin^{2}(u)du$.

Step2: Use double - angle formula

Recall that $\sin^{2}(u)=\frac{1 - \cos(2u)}{2}$. So $\frac{9}{7}\int\sin^{2}(u)du=\frac{9}{7}\int\frac{1-\cos(2u)}{2}du=\frac{9}{14}\int(1 - \cos(2u))du$.

Step3: Integrate term - by - term

$\frac{9}{14}\int(1 - \cos(2u))du=\frac{9}{14}\left(\int 1du-\int\cos(2u)du\right)$. The integral of $1$ with respect to $u$ is $u$, and for $\int\cos(2u)du$, let $v = 2u$, $dv = 2du$, then $\int\cos(2u)du=\frac{1}{2}\sin(2u)$. So $\frac{9}{14}\left(u-\frac{1}{2}\sin(2u)\right)+C$.

Step4: Substitute back $u = x^{7}$

We get $\frac{9}{14}\left(x^{7}-\frac{1}{2}\sin(2x^{7})\right)+C=\frac{9x^{7}}{14}-\frac{9\sin(2x^{7})}{28}+C$.

Answer:

$\frac{9x^{7}}{14}-\frac{9\sin(2x^{7})}{28}$