Evaluate the integral.
-$$\int - 8 x \cos 2 x d x$$
A) $- 2 \cos 2 x - 4 \sin 2 x + C$
B) $- 2 \cos 2 x - 4 x \sin 8 x + C$
C) $- 2 \cos 2 x - 4 x \sin 2 x + C$
D) $- 4 \cos 2 x - 8 x \sin 2 x + C$

Question

Quizplus · Accepted Answer

This integral can be solved using integration by parts, where $u = x$ and $dv = -8\cos(2x)dx$. The derivative of $u$ is $du = dx$, and the integral of $dv$ is $v = -4\sin(2x)$. Applying the integration by parts formula $\int u dv = uv - \int v du$, we get $-4x\sin(2x) + \int 4\sin(2x)dx$. The integral of $4\sin(2x)$ is $-2\cos(2x)$, leading to the final answer of $-4x\sin(2x) - 2\cos(2x) + C$.

Evaluate the Integral ∫−8xcos⁡2xdx\int - 8 x \cos 2 x d x∫−8xcos2xdx

Evaluate the Integral $\int - 8 x \cos 2 x d x$