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The integral of \( x \cos(2x) \) with respect to \( x \) is \( \frac{1}{2}x\sin(2x) + \frac{1}{4}\cos(2x) + C \).
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The Deep Dive
To solve the integral \( \int x \cos 2x \, dx \), we can use integration by parts. Let \( u = x \) and \( dv = \cos 2x \, dx \). Then, \( du = dx \) and \( v = \frac{1}{2} \sin 2x \). Now, applying integration by parts yields: \[ \int x \cos 2x \, dx = uv - \int v \, du = x \left(\frac{1}{2} \sin 2x\right) - \int \left(\frac{1}{2} \sin 2x\right) \, dx. \] Calculating the remaining integral, we have: \[ \int \left(\frac{1}{2} \sin 2x\right) dx = -\frac{1}{4} \cos 2x + C. \] Combining everything, we get: \[ \int x \cos 2x \, dx = \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x + C. \]
