Evaluate the integral.
-$$\int - 6 x \cos 9 x d x$$
A) $- \frac { 2 } { 3 } \cos 9 x - 6 x \sin 9 x + C$
B) $- \frac { 2 } { 27 } \cos 9 x - \frac { 2 } { 3 } x \sin 9 x + C$
C) $- \frac { 2 } { 27 } \cos 9 x - \frac { 2 } { 3 } x \sin 6 x + C$
D) $- \frac { 2 } { 27 } \cos 9 x - \frac { 2 } { 3 } \sin 9 x + C$

Question

Quizplus · Accepted Answer

This integral requires the use of integration by parts, where $u = x$ and $dv = -6\cos(9x)dx$. After integrating $dv$ and differentiating $u$, we get $du = dx$ and $v = -\frac{2}{3}\sin(9x)$. Applying the integration by parts formula $\int u dv = uv - \int v du$, we obtain $-\frac{2}{3}x\sin(9x) - \int -\frac{2}{3}\sin(9x)dx$. Integrating the second term gives $-\frac{2}{27}\cos(9x)$, leading to the final answer $-\frac{2}{27}\cos(9x) - \frac{2}{3}x\sin(9x) + C$.

Evaluate the Integral ∫−6xcos⁡9xdx\int - 6 x \cos 9 x d x∫−6xcos9xdx

Evaluate the Integral $\int - 6 x \cos 9 x d x$