\\(\\int_{-3}^{-7} sin^{3}(6x)dx\\)

Explanation:

Step1: Rewrite $\sin^{3}(6x)$

Use $\sin^{3}u=\sin u(1 - \cos^{2}u)$, so $\sin^{3}(6x)=\sin(6x)(1 - \cos^{2}(6x))$. Then the integral becomes $\int_{-3}^{-7}\sin(6x)(1 - \cos^{2}(6x))dx$.

Step2: Use substitution

Let $u = \cos(6x)$, then $du=-6\sin(6x)dx$ and $\sin(6x)dx=-\frac{1}{6}du$. When $x=-3$, $u=\cos(- 18)$; when $x = - 7$, $u=\cos(-42)$. The integral is $-\frac{1}{6}\int_{\cos(-18)}^{\cos(-42)}(1 - u^{2})du$.

Step3: Integrate term - by - term

Integrate $1 - u^{2}$ with respect to $u$. $\int(1 - u^{2})du=u-\frac{1}{3}u^{3}+C$. Then $-\frac{1}{6}\left[u-\frac{1}{3}u^{3}\right]_{\cos(-18)}^{\cos(-42)}$.

Step4: Evaluate the definite integral

$-\frac{1}{6}\left[\left(\cos(-42)-\frac{1}{3}\cos^{3}(-42)\right)-\left(\cos(-18)-\frac{1}{3}\cos^{3}(-18)\right)\right]$. Since $\cos(-\alpha)=\cos\alpha$, we have $-\frac{1}{6}\left[\left(\cos(42)-\frac{1}{3}\cos^{3}(42)\right)-\left(\cos(18)-\frac{1}{3}\cos^{3}(18)\right)\right]$.

Answer:

$-\frac{1}{6}\left[\left(\cos(42)-\frac{1}{3}\cos^{3}(42)\right)-\left(\cos(18)-\frac{1}{3}\cos^{3}(18)\right)\right]$