Evaluate-x-cos-2-x-dx-

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Question Number 40 by user3 last updated on 25/Jan/15

$$\mathrm{Evaluate}\int x\cos {}^{2}xdx.$$

Answered by user3 last updated on 03/Nov/14

$$\int x{\cos }^{2}xdx=\int x\left(\frac{1+\cos 2x}{2}\right)dx$$$$=\frac{1}{2}\int xdx+\frac{1}{2}\int x\cos 2xdx$$$$=\frac{x^2}{4}+\frac{1}{2}\cdot [x\cdot \int \cos 2xdx-\int \{\frac{d}{dx}\left(x\right)\cdot \int \cos 2xdx\}dx]$$$$=\frac{x^2}{4}+\frac{1}{2}[\frac{x\sin 2x}{2}-\int \frac{\sin 2x}{2}dx]$$$$=\frac{x^2}{4}+\frac{x\sin 2x}{4}-\frac{1}{4}\times \frac{(-\cos 2x)}{2}+C$$$$=\frac{x^2}{4}+\frac{x\sin 2x}{4}+\frac{\cos 2x}{8}+C$$

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