Integration by Parts

Reversing the product rule: ∫u dv = uv − ∫v du. The standard technique for integrating products of functions.

Integration by parts is the product rule run backward

d(uv) = u dv + v du

rearrange

∫ u dv = uv - ∫ v du

Trade the hard part

Choose u so differentiating makes it simpler, and choose dv so integrating is manageable.

Definition

Integration by Parts is the integral analogue of the product rule. For differentiable functions uu and vv:

∫u dv=uv−∫v du\int u\, dv = uv - \int v\, du

How to use it:

  1. Choose uu and dvdv from your integrand.
  2. Compute dudu (differentiate uu) and vv (integrate dvdv).
  3. Apply the formula: ∫u dv=uv−∫v du\int u\, dv = uv - \int v\, du.
  4. Hope that ∫v du\int v\, du is easier than the original integral.

LIATE rule for choosing uu (pick whichever comes first):

  • Logarithms
  • Inverse trig
  • Algebraic (polynomials)
  • Trig
  • Exponentials
∫x·eˣ dx

Let u=xu = x, dv=ex dxdv = e^x\, dx. Then du=dxdu = dx, v=exv = e^x.

∫xex dx=xex−∫ex dx=xex−ex+C=ex(x−1)+C\int x e^x\, dx = x e^x - \int e^x\, dx = x e^x - e^x + C = e^x(x - 1) + C

Check: ddx[ex(x−1)]=ex(x−1)+ex=xex\frac{d}{dx}[e^x(x-1)] = e^x(x-1) + e^x = xe^x ✓

Try it

Evaluate ∫xcos⁡x dx\int x\cos x\, dx.

Solution

Let u=xu = x, dv=cos⁡x dxdv = \cos x\, dx. Then du=dxdu = dx, v=sin⁡xv = \sin x.

∫xcos⁡x dx=xsin⁡x−∫sin⁡x dx=xsin⁡x+cos⁡x+C\int x\cos x\, dx = x\sin x - \int \sin x\, dx = x\sin x + \cos x + C

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