Find $\int x \cos 2 x d x$ .
A) $\frac{1}{2} x \sin 2 x+\frac{1}{4} \cos 2 x+C$
B) $\frac{1}{2} x \sin 2 x-\frac{1}{2} \cos 2 x+C$
C) $-\frac{x^{2}}{8} \sin 2 x+C$
D) $\frac{x^{2}}{4} \sin 2 x+C$

Question

Quizplus · Accepted Answer

This integral can be solved using integration by parts, where $u = x$ and $dv = \cos 2x dx$. The formula for integration by parts is $\int u dv = uv - \int v du$. Applying this, we get $\frac{1}{2} x \sin 2 x+\frac{1}{4} \cos 2 x+C$ as the correct answer.

Find ∫xcos⁡2xdx\int x \cos 2 x d x∫xcos2xdx A) 12xsin⁡2x+14cos⁡2x+C\frac{1}{2} x \sin 2 x+\frac{1}{4} \cos 2 x+C21​xsin2x+41​cos2x+C

Find $\int x \cos 2 x d x$
A) $\frac{1}{2} x \sin 2 x+\frac{1}{4} \cos 2 x+C$