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What is the Integral of $x \cos2x$?
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The integration is of the form
$I=\int x \cos 2 x d x$
Here the first function is $x$ and the second function is $\cos 2 x$
$I=\int x \cos 2 x d x-\cdots \text { (i) }$
Using the formula for integration by parts, we have
$\int[f(x) g(x)] d x=f(x) \int g(x) d x-\int\left[\frac{d}{d x} f(x) \int g(x) d x\right] d x$
Using the formula above, equation (i) becomes
$I=x \int \cos 2 x d x-\int\left[\frac{d}{d x} x\left(\int \cos 2 x d x\right)\right] d x$
Using the integral rule $\int \cos k x d x=\frac{\sin k x}{k}+c$, we have
\begin{aligned} &I=x \frac{\sin 2 x}{2}-\int\left[(1) \frac{\sin 2 x}{2}\right] d x \\ &\Rightarrow I=\frac{x}{2} \sin 2 x-\frac{1}{2} \int \sin 2 x d x \end{aligned}
Using the integral rule $\int \sin k x d x=-\frac{\cos k x}{k}+c$, we have
$\begin{gathered} I=\frac{x}{2} \sin 2 x-\frac{1}{2}\left(\frac{-\cos 2 x}{2}\right)+c \\ \Rightarrow I=\frac{x}{2} \sin 2 x+\frac{1}{4} \cos 2 x+c \\ \Rightarrow \int x \cos 2 x d x=\frac{x}{2} \sin 2 x+\frac{1}{4} \cos 2 x+c \end{gathered}$
by Diamond (38,951 points)

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