Topic 6: Integration by Parts

Justification of Integration by Substitution

  • Deals with integrals of the form: ∫f(g(x))⋅g′(x)dx\int f(g(x)) \cdot g'(x) dx, where we know how to integrate ff.
  • Let ∫f(x)dx=F(x)\int f(x) dx = F(x), which means F′(x)=f(x)F'(x) = f(x).
  • Consider ddx[F(g(x))]\frac{d}{dx} [F(g(x))]. Using the chain rule, this is F′(g(x))⋅g′(x)F'(g(x)) \cdot g'(x).
  • Since F′(x)=f(x)F'(x) = f(x), we have f(g(x))⋅g′(x)f(g(x)) \cdot g'(x).
  • Therefore, ∫f(g(x))⋅g′(x)dx=F(g(x))+C\int f(g(x)) \cdot g'(x) dx = F(g(x)) + C.
  • It revolves around the chain rule.
  • You're integrating the outer function to get capital FF and we do that with respect to U.
  • The last step is to substitute back in for u to get it in terms of x.

Integration by Parts

  • Integration by parts is like doing the product rule in reverse.
  • Product Rule: ddx(uv)=u⋅v′+v⋅u′\frac{d}{dx}(uv) = u \cdot v' + v \cdot u'
  • Integrate both sides: ∫ddx(uv)dx=∫(u⋅v′+v⋅u′)dx\int \frac{d}{dx}(uv) dx = \int (u \cdot v' + v \cdot u') dx
  • Which simplifies to: uv=∫udvdxdx+∫vdudxdxuv = \int u \frac{dv}{dx} dx + \int v \frac{du}{dx} dx
  • Rearranging to get the integration by parts formula:
    ∫udvdxdx=uv−∫vdudxdx\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx

Example 4A: Integral of x⋅exx \cdot e^x

  • Integral: ∫x⋅exdx\int x \cdot e^x dx
  • Identify uu and dvdx\frac{dv}{dx}: Let u=xu = x and dvdx=ex\frac{dv}{dx} = e^x.
  • Find dudx\frac{du}{dx} and vv: dudx=1\frac{du}{dx} = 1 and v=exv = e^x.
  • Apply the integration by parts formula: ∫udvdxdx=uv−∫vdudxdx\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx
  • Substitute: ∫x⋅exdx=x⋅ex−∫ex⋅1dx\int x \cdot e^x dx = x \cdot e^x - \int e^x \cdot 1 dx
  • Evaluate the remaining integral: x⋅ex−ex+Cx \cdot e^x - e^x + C.

Alternative (Incorrect) Approach

  • Let u=exu = e^x and dvdx=x\frac{dv}{dx} = x.
  • Then, dudx=ex\frac{du}{dx} = e^x and v=x22v = \frac{x^2}{2}.
  • Applying integration by parts: ∫x⋅exdx=ex⋅x22−∫x22⋅exdx\int x \cdot e^x dx = e^x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot e^x dx
  • The new integral, ∫x22⋅exdx\int \frac{x^2}{2} \cdot e^x dx, is more complex than the original.

Choosing uu and dvdx\frac{dv}{dx}

  • Prioritize choosing uu in the following order:
    1. Logarithms
    2. Power of xx
    3. Exponential Functions
    4. Trigonometric Functions
  • The goal is to choose uu such that dudx\frac{du}{dx} is simpler than uu.
  • In the example, choosing u=xu = x works well because dudx=1\frac{du}{dx} = 1 (simpler).
  • Choosing u=exu = e^x doesn't simplify the integral because dudx=ex\frac{du}{dx} = e^x.

Example 4B: Integral of x2⋅e−2xx^2 \cdot e^{-2x}

  • Integral: ∫x2⋅e−2xdx\int x^2 \cdot e^{-2x} dx
  • Let u=x2u = x^2 and dvdx=e−2x\frac{dv}{dx} = e^{-2x}.
  • Then, dudx=2x\frac{du}{dx} = 2x and v=−12e−2xv = -\frac{1}{2} e^{-2x}.
  • Apply integration by parts: ∫x2⋅e−2xdx=x2⋅(−12e−2x)−∫(−12e−2x)⋅2xdx\int x^2 \cdot e^{-2x} dx = x^2 \cdot (-\frac{1}{2} e^{-2x}) - \int (-\frac{1}{2} e^{-2x}) \cdot 2x dx
  • Simplify: −12x2e−2x+∫x⋅e−2xdx-\frac{1}{2} x^2 e^{-2x} + \int x \cdot e^{-2x} dx
  • Apply integration by parts again to the integral ∫x⋅e−2xdx\int x \cdot e^{-2x} dx:
    • Let u=xu = x and dvdx=e−2x\frac{dv}{dx} = e^{-2x}.
    • Then, dudx=1\frac{du}{dx} = 1 and v=−12e−2xv = -\frac{1}{2} e^{-2x}.
  • ∫x⋅e−2xdx=x⋅(−12e−2x)−∫(−12e−2x)⋅1dx\int x \cdot e^{-2x} dx = x \cdot (-\frac{1}{2} e^{-2x}) - \int (-\frac{1}{2} e^{-2x}) \cdot 1 dx
  • ∫x⋅e−2xdx=−12xe−2x+12∫e−2xdx\int x \cdot e^{-2x} dx = -\frac{1}{2} x e^{-2x} + \frac{1}{2} \int e^{-2x} dx
  • ∫x⋅e−2xdx=−12xe−2x−14e−2x+C\int x \cdot e^{-2x} dx = -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} + C
  • Substitute back into the original equation:
    • ∫x2⋅e−2xdx=−12x2e−2x−12xe−2x−14e−2x+C\int x^2 \cdot e^{-2x} dx = -\frac{1}{2} x^2 e^{-2x} -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} + C

Example 4C: Integral of ln⁡(x)\ln(x)

  • Integral: ∫ln⁡(x)dx\int \ln(x) dx
  • Rewrite as a product: ∫ln⁡(x)⋅1dx\int \ln(x) \cdot 1 dx
  • Let u=ln⁡(x)u = \ln(x) and dvdx=1\frac{dv}{dx} = 1.
  • Then, dudx=1x\frac{du}{dx} = \frac{1}{x} and v=xv = x.
  • Apply integration by parts: ∫ln⁡(x)⋅1dx=x⋅ln⁡(x)−∫x⋅1xdx\int \ln(x) \cdot 1 dx = x \cdot \ln(x) - \int x \cdot \frac{1}{x} dx
  • Simplify: x⋅ln⁡(x)−∫1dxx \cdot \ln(x) - \int 1 dx
  • x⋅ln⁡(x)−x+Cx \cdot \ln(x) - x + C

Example 4D: Integral of ex⋅sin⁡(x)e^x \cdot \sin(x)

  • Integral: ∫ex⋅sin⁡(x)dx\int e^x \cdot \sin(x) dx
  • Let u=exu = e^x and dvdx=sin⁡(x)\frac{dv}{dx} = \sin(x).
  • Then, dudx=ex\frac{du}{dx} = e^x and v=−cos⁡(x)v = -\cos(x).
  • Apply integration by parts: ∫ex⋅sin⁡(x)dx=ex⋅(−cos⁡(x))−∫(−cos⁡(x))⋅exdx\int e^x \cdot \sin(x) dx = e^x \cdot (-\cos(x)) - \int (-\cos(x)) \cdot e^x dx
  • Simplify: −ex⋅cos⁡(x)+∫ex⋅cos⁡(x)dx-e^x \cdot \cos(x) + \int e^x \cdot \cos(x) dx
  • Apply integration by parts again to the integral ∫ex⋅cos⁡(x)dx\int e^x \cdot \cos(x) dx:
    • Let u=exu = e^x and dvdx=cos⁡(x)\frac{dv}{dx} = \cos(x).
    • Then, dudx=ex\frac{du}{dx} = e^x and v=sin⁡(x)v = \sin(x).
  • ∫ex⋅cos⁡(x)dx=ex⋅sin⁡(x)−∫sin⁡(x)⋅exdx\int e^x \cdot \cos(x) dx = e^x \cdot \sin(x) - \int \sin(x) \cdot e^x dx
  • Substitute back into the original equation:
    • ∫ex⋅sin⁡(x)dx=−ex⋅cos⁡(x)+ex⋅sin⁡(x)−∫sin⁡(x)⋅exdx\int e^x \cdot \sin(x) dx = -e^x \cdot \cos(x) + e^x \cdot \sin(x) - \int \sin(x) \cdot e^x dx
  • Rearrange the equation:
    • 2∫ex⋅sin⁡(x)dx=−ex⋅cos⁡(x)+ex⋅sin⁡(x)2 \int e^x \cdot \sin(x) dx = -e^x \cdot \cos(x) + e^x \cdot \sin(x)
  • Divide by 2:
    • ∫ex⋅sin⁡(x)dx=12ex(sin⁡(x)−cos⁡(x))+C\int e^x \cdot \sin(x) dx = \frac{1}{2} e^x (\sin(x) - \cos(x)) + C