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3-2 (b) Integration by substitution

3-2 (b) Integration by substitution

Point 4 of 12 — Lesson 1: Integration Techniques (Substitution and By Parts) and Definite IntegralsQuestion bank: 9 questions (9 with pictures)

04 3-2 (b) Integration by substitution

Idea

  • Used for integrals of the form ∫f(g(x)) g′(x) dx.
  • If z=g(x) then dz=g′(x) dx, so ∫f(g(x)) g′(x) dx=∫f(z) dz
  • Steps: choose z; find dz; rewrite the whole integral in z (including x and dx); integrate; return to x.
  • Choose z so that the new integral is simpler and a standard one; if the first choice fails, try another.
✍️ Worked examples

Example 6: Find (a) ∫x3(2x4−7)5 dx (b) ∫x+4(x2+8x)3 dx.

Solution:

  • (a) Put z=2x4−7, so dz=8x3 dx.
  • ∫z5⋅dz8=18⋅z66+c.
  • =(2x4−7)648+c.
  • (b) Put z=x2+8x, so dz=(2x+8) dx=2(x+4) dx.
  • ∫x+4z3 dx=12∫z−3 dz=12⋅z−2−2+c.
  • =−14(x2+8x)2+c.

When the integrand has an extra factor of x

  • Put z equal to the inner expression and write x in terms of z too.
✍️ Worked examples

Example 7: Find (a) ∫x(x+4)5 dx (b) ∫xx−1 dx.

Solution:

  • (a) Put z=x+4, so x=z−4 and dx=dz.
  • ∫(z−4)z5 dz=∫(z6−4z5) dz=z77−2z63+c.
  • =z6(3z−14)21+c=(x+4)6(3x−2)21+c.
  • (b) Put z2=x−1, so x=z2+1 and dx=2z dz.
  • ∫(z2+1) z⋅2z dz=2∫(z4+z2) dz=2z55+2z33+c.
  • =215z3(3z2+5)+c=215(x−1)3/2(3x+2)+c.

Two rules proved by substitution

  • Put z=f(x) in ∫[f(x)]nf′(x) dx: it becomes ∫zn dz=zn+1n+1+c, so ∫[f(x)]nf′(x) dx=[f(x)]n+1n+1+c(n≠−1)
  • Put z=f(x) in ∫f′(x)f(x) dx: it becomes ∫dzz=ln⁡∣z∣+c.

📝 Question bank for this point

9 questions linked to this point only, 9 of them with pictures.