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Question

Question: Evaluate: $\int x \sin 2x dx$...

Evaluate: ∫xsin⁡2xdx\int x \sin 2x dx

A

xcos⁡2x2+sin⁡2x4+C\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} + C

B

xcos⁡2x2−sin⁡2x4+C\frac{x \cos 2x}{2} - \frac{\sin 2x}{4} + C

C

−xcos⁡2x2+sin⁡2x4+C-\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} + C

D

xcos⁡2x4+sin⁡2x2+C\frac{x \cos 2x}{4} + \frac{\sin 2x}{2} + C

Answer

−xcos⁡2x2+sin⁡2x4+C-\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} + C

Explanation

Solution

The integral ∫xsin⁡2xdx\int x \sin 2x dx is solved using integration by parts, ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

  1. Choose u=xu=x (algebraic) and dv=sin⁡2x dxdv=\sin 2x \, dx (trigonometric) based on the LIATE rule.
  2. This gives du=dxdu=dx and v=∫sin⁡2x dx=−12cos⁡2xv=\int \sin 2x \, dx = -\frac{1}{2}\cos 2x.
  3. Substitute these into the formula: ∫xsin⁡2x dx=x(−12cos⁡2x)−∫(−12cos⁡2x)dx\int x \sin 2x \, dx = x \left(-\frac{1}{2}\cos 2x\right) - \int \left(-\frac{1}{2}\cos 2x\right) dx =−xcos⁡2x2+12∫cos⁡2x dx= -\frac{x\cos 2x}{2} + \frac{1}{2} \int \cos 2x \, dx
  4. The remaining integral ∫cos⁡2x dx=12sin⁡2x\int \cos 2x \, dx = \frac{1}{2}\sin 2x.
  5. Substitute this back: =−xcos⁡2x2+12(12sin⁡2x)+C= -\frac{x\cos 2x}{2} + \frac{1}{2}\left(\frac{1}{2}\sin 2x\right) + C =−xcos⁡2x2+sin⁡2x4+C= -\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C.