evaluate.\n\\(\\int_{-\\frac{\\pi}{4}}^{\\frac{\\pi}{6}}(2x + 4)\\sin(6x)dx=\\)

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = 2x + 4$ and $dv=\sin(6x)dx$. Then $du = 2dx$ and $v=-\frac{1}{6}\cos(6x)$.

Step2: Apply the integration - by - parts formula

$\int(2x + 4)\sin(6x)dx=-\frac{1}{6}(2x + 4)\cos(6x)+\frac{1}{3}\int\cos(6x)dx$.

Step3: Integrate $\int\cos(6x)dx$

$\int\cos(6x)dx=\frac{1}{6}\sin(6x)+C$. So $\int(2x + 4)\sin(6x)dx=-\frac{1}{6}(2x + 4)\cos(6x)+\frac{1}{18}\sin(6x)+C$.

Step4: Evaluate the definite integral

$\left[-\frac{1}{6}(2x + 4)\cos(6x)+\frac{1}{18}\sin(6x)\right]_{-\frac{\pi}{4}}^{\frac{\pi}{6}}$ First, substitute $x=\frac{\pi}{6}$: $-\frac{1}{6}(2\times\frac{\pi}{6}+4)\cos(\pi)+\frac{1}{18}\sin(\pi)=-\frac{1}{6}(\frac{\pi}{3}+4)(- 1)+0=\frac{\pi + 12}{18}$. Then, substitute $x =-\frac{\pi}{4}$: $-\frac{1}{6}(2\times(-\frac{\pi}{4})+4)\cos(-\frac{3\pi}{2})+\frac{1}{18}\sin(-\frac{3\pi}{2})$ $=0+\frac{1}{18}(1)=\frac{1}{18}$.

Step5: Subtract the two results

$\frac{\pi + 12}{18}-\frac{1}{18}=\frac{\pi + 11}{18}$.

Answer:

$\frac{\pi + 11}{18}$