evaluate.\n int_{-\frac{pi}{6}}^{\frac{pi}{4}} (4x - 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 $u = 4x - 4$ and $dv=\sin(6x)dx$. Then $du = 4dx$ and $v=-\frac{1}{6}\cos(6x)$. [ \begin{align*} \int(4x - 4)\sin(6x)dx&=(4x - 4)\left(-\frac{1}{6}\cos(6x)\right)-\int\left(-\frac{1}{6}\cos(6x)\right)\times4dx\ &=-\frac{4x - 4}{6}\cos(6x)+\frac{2}{3}\int\cos(6x)dx\ &=-\frac{4x - 4}{6}\cos(6x)+\frac{2}{3}\times\frac{1}{6}\sin(6x)+C\ &=-\frac{4x - 4}{6}\cos(6x)+\frac{1}{9}\sin(6x)+C \end{align*} ]

Step2: Evaluate the definite integral

[ \begin{align*} &\left[-\frac{4x - 4}{6}\cos(6x)+\frac{1}{9}\sin(6x)\right]_{-\frac{\pi}{6}}^{\frac{\pi}{4}}\ =&\left(-\frac{4\times\frac{\pi}{4}-4}{6}\cos\left(6\times\frac{\pi}{4}\right)+\frac{1}{9}\sin\left(6\times\frac{\pi}{4}\right)\right)-\left(-\frac{4\times\left(-\frac{\pi}{6}\right)-4}{6}\cos\left(6\times\left(-\frac{\pi}{6}\right)\right)+\frac{1}{9}\sin\left(6\times\left(-\frac{\pi}{6}\right)\right)\right)\ =&\left(-\frac{\pi - 4}{6}\cos\left(\frac{3\pi}{2}\right)+\frac{1}{9}\sin\left(\frac{3\pi}{2}\right)\right)-\left(-\frac{-\frac{2\pi}{3}-4}{6}\cos(-\pi)+\frac{1}{9}\sin(-\pi)\right)\ =&\left(0-\frac{1}{9}\right)-\left(-\frac{-\frac{2\pi + 12}{3}}{6}\times(-1)+0\right)\ =&-\frac{1}{9}-\frac{2\pi + 12}{18}\ =&-\frac{2 + 2\pi+12}{18}\ =&-\frac{2\pi + 14}{18}\ =&-\frac{\pi+7}{9} \end{align*} ]

Answer:

$-\frac{\pi + 7}{9}$