Integration by Parts

Integrate the product rule and get ∫u dv=[uv]−∫v du\int u\,dv = [uv] - \int v\,du — split the area into two pieces and compute each. Move bb: both sides agree to the last decimal.

∫abu dv=[uv]ab−∫abv du\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du
∫xex dx=(x−1)ex+C\int x e^x\,dx = (x-1)e^x + C

Examples & Exercises

Prove integration by parts completely: if u,vu, v are C1C^1 on [a,b][a,b], then ∫abu dv=[uv]ab−∫abv du\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du. Integrate the product rule (uv)′=u′v+uv′(uv)' = u'v + uv' and apply the fundamental theorem of calculus.

∫abu(x) v′(x) dx=[u(x) v(x)]ab−∫abv(x) u′(x) dx\int_a^b u(x)\,v'(x)\,dx = \bigl[u(x)\,v(x)\bigr]_a^b - \int_a^b v(x)\,u'(x)\,dx
u=xu = x, dv=ex dxdv = e^x\,dx: LHS ≈8.389056099\approx 8.389056099, and [uv]02−∫02v du≈2e2−(e2−1)[uv]_0^2 - \int_0^2 v\,du \approx 2e^2 - (e^2 - 1) matches — the top area equals [uv]0b[uv]_0^b minus the bottom area.
Show proof
  1. Since u,vu, v are C1C^1, u′,v′u', v' are continuous, the product uvuv is C1C^1, and (uv)′=u′v+uv′(uv)' = u'v + uv' (the product rule). Each term u′vu'v, uv′uv' is a product of continuous functions, hence continuous — so both are integrable on [a,b][a,b].
  2. Integrate (uv)′=u′v+uv′(uv)' = u'v + uv' over [a,b][a,b]: ∫ab(uv)′ dx=∫abu′v dx+∫abuv′ dx\int_a^b (uv)'\,dx = \int_a^b u'v\,dx + \int_a^b uv'\,dx. The FTC turns the left side into ∫ab(uv)′ dx=[uv]ab\int_a^b (uv)'\,dx = [uv]_a^b.
  3. Hence [uv]ab=∫abu′v dx+∫abuv′ dx[uv]_a^b = \int_a^b u'v\,dx + \int_a^b uv'\,dx, and moving ∫abu′v dx\int_a^b u'v\,dx across gives ∫abuv′ dx=[uv]ab−∫abu′v dx\int_a^b uv'\,dx = [uv]_a^b - \int_a^b u'v\,dx — i.e. ∫abu dv=[uv]ab−∫abv du\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du. Complete. (Tao Proposition 11.10.1.)

Compute ∫xex dx=(x−1)ex+C\int x e^x\,dx = (x-1)e^x + C completely via integration by parts, then differentiate the answer to verify.

∫xex dx=(x−1)ex+C,ddx[(x−1)ex]=xex\int x e^x\,dx = (x-1)e^x + C, \qquad \frac{d}{dx}\bigl[(x-1)e^x\bigr] = x e^x
b=2b = 2: ∫02xex dx=(2−1)e2+1=e2+1≈8.389056099\int_0^2 x e^x\,dx = (2-1)e^2 + 1 = e^2 + 1 \approx 8.389056099 — matches the scene's closed form.
Show proof
  1. Set u=xu = x, dv=ex dxdv = e^x\,dx; then du=dxdu = dx, v=exv = e^x. The formula gives ∫xex dx=xex−∫ex dx=xex−ex+C=(x−1)ex+C\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = (x-1)e^x + C.
  2. Verification (differentiate back): ddx[(x−1)ex]=1⋅ex+(x−1)ex=xex\frac{d}{dx}[(x-1)e^x] = 1\cdot e^x + (x-1)e^x = x e^x (product rule) — matches the original integrand, so the antiderivative is correct. As a definite integral, ∫02xex dx=[(x−1)ex]02=e2−(−1)=e2+1≈8.389056099\int_0^2 x e^x\,dx = [(x-1)e^x]_0^2 = e^2 - (-1) = e^2 + 1 \approx 8.389056099.

Derive completely via integration by parts that In=∫0π/2sin⁡nx dxI_n = \int_0^{\pi/2} \sin^n x\,dx satisfies In=n−1n In−2I_n = \frac{n-1}{n}\,I_{n-2} (use u=sin⁡n−1xu = \sin^{n-1}x, dv=sin⁡x dxdv = \sin x\,dx). Then compute I0=π2I_0 = \frac{\pi}{2} and I1=1I_1 = 1.

In=n−1n In−2,I0=π2,I1=1I_n = \frac{n-1}{n}\,I_{n-2}, \qquad I_0 = \frac{\pi}{2}, \quad I_1 = 1
I2=π4,I3=23,I4=3π16I_2 = \frac{\pi}{4}, \quad I_3 = \frac{2}{3}, \quad I_4 = \frac{3\pi}{16}
The trigonometric pair: u=xu = x, dv=sin⁡x dxdv = \sin x\,dx gives v=−cos⁡xv = -\cos x and ∫0bxsin⁡x dx=−bcos⁡b+sin⁡b\int_0^b x\sin x\,dx = -b\cos b + \sin b — the same parts structure produces the reduction formula.
Show proof
  1. For n≥2n \ge 2, write In=∫0π/2sin⁡n−1x⋅sin⁡x dxI_n = \int_0^{\pi/2} \sin^{n-1}x \cdot \sin x\,dx and set u=sin⁡n−1xu = \sin^{n-1}x, dv=sin⁡x dxdv = \sin x\,dx: then du=(n−1)sin⁡n−2xcos⁡x dxdu = (n-1)\sin^{n-2}x\cos x\,dx, v=−cos⁡xv = -\cos x. Parts gives In=[−cos⁡x sin⁡n−1x]0π/2+(n−1)∫0π/2sin⁡n−2x cos⁡2x dxI_n = [-\cos x\,\sin^{n-1}x]_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2}x\,\cos^2 x\,dx.
  2. The boundary term vanishes: cos⁡=0\cos = 0 at x=π/2x = \pi/2 and sin⁡=0\sin = 0 at x=0x = 0 (since n−1≥1n - 1 \ge 1). With cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x this becomes In=(n−1)∫0π/2(sin⁡n−2x−sin⁡nx) dx=(n−1)(In−2−In)I_n = (n-1)\int_0^{\pi/2} (\sin^{n-2}x - \sin^n x)\,dx = (n-1)(I_{n-2} - I_n). Rearranging, n In=(n−1) In−2n\,I_n = (n-1)\,I_{n-2}, i.e. In=n−1n In−2I_n = \frac{n-1}{n}\,I_{n-2}.
  3. Base cases: I0=∫0π/21 dx=π2I_0 = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}, I1=∫0π/2sin⁡x dx=[−cos⁡x]0π/2=1I_1 = \int_0^{\pi/2} \sin x\,dx = [-\cos x]_0^{\pi/2} = 1. The recurrence gives I2=12⋅π2=π4I_2 = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}, I3=23⋅1=23I_3 = \frac{2}{3}\cdot 1 = \frac{2}{3}, I4=34⋅π4=3π16I_4 = \frac{3}{4}\cdot\frac{\pi}{4} = \frac{3\pi}{16} — the starting point of the Wallis integrals.

10. Integration