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(b) Use integration by parts to evaluate ∫0π2x⋅cos⁡x dx \int_0^{\frac{\pi}{2}} x \cdot \cos x \, dx

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Q. (b) Use integration by parts to evaluate ∫0π2x⋅cos⁡x dx \int_0^{\frac{\pi}{2}} x \cdot \cos x \, dx
  1. Set up integration formula: Step 11: Set up the integration by parts formula.\newlineWe use the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. Let u=xu = x and dv=cos⁡(x) dxdv = \cos(x) \, dx.\newlineThen, du=dxdu = dx and v=∫cos⁡(x) dx=sin⁡(x)v = \int\cos(x) \, dx = \sin(x).
  2. Apply integration by parts: Step 22: Apply the integration by parts formula.\newlinePlug in uu, vv, dudu, and dvdv into the formula:\newline∫xcos⁡(x) dx=xsin⁡(x)−∫sin⁡(x) dx\int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx.
  3. Integrate sin⁡(x)\sin(x): Step 33: Integrate ∫sin⁡(x) dx\int \sin(x) \, dx. The integral of sin⁡(x)\sin(x) is −cos⁡(x)-\cos(x), so: ∫xcos⁡(x) dx=xsin⁡(x)+cos⁡(x)\int x \cos(x) \, dx = x \sin(x) + \cos(x).
  4. Evaluate definite integral: Step 44: Evaluate the definite integral from 00 to π/2\pi/2.\newlinePlug in the limits of integration:\newline[xsin⁡(x)+cos⁡(x)][x \sin(x) + \cos(x)] from 00 to π/2\pi/2.\newline= (π/2⋅sin⁡(π/2)+cos⁡(π/2))−(0⋅sin⁡(0)+cos⁡(0))(\pi/2 \cdot \sin(\pi/2) + \cos(\pi/2)) - (0 \cdot \sin(0) + \cos(0))\newline= (π/2⋅1+0)−(0⋅0+1)(\pi/2 \cdot 1 + 0) - (0 \cdot 0 + 1)\newline= π/2−1\pi/2 - 1.

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