$\int\left(x^{2}+2\right) \cos 2 x d x=\frac{1}{2}\left(x^{2}+2\right) \sin 2 x-1 x \cos 2 x-\frac{1}{4} \sin 2 x+C$ .

Question

Quizplus · Accepted Answer

The correct integral of $\int\left(x^{2}+2\right) \cos 2 x d x$ involves applying integration by parts and possibly substitution, but the given solution has a mistake in the coefficient of the $x \cos 2x$ term. It should be $\frac{x}{2}$ instead of $1x$, and the constant of integration $C$ is correctly placed, but the overall structure of the solution suggests a misapplication of the integration by parts formula.

∫(x2+2)cos⁡2xdx=12(x2+2)sin⁡2x−1xcos⁡2x−14sin⁡2x+C\int\left(x^{2}+2\right) \cos 2 x d x=\frac{1}{2}\left(x^{2}+2\right) \sin 2 x-1 x \cos 2 x-\frac{1}{4} \sin 2 x+C∫(x2+2)cos2xdx=21​(x2+2)sin2x−1xcos2x−41​sin2x+C

$\int\left(x^{2}+2\right) \cos 2 x d x=\frac{1}{2}\left(x^{2}+2\right) \sin 2 x-1 x \cos 2 x-\frac{1}{4} \sin 2 x+C$