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2.2: Trigonometric Substitution

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When we have integrals that involve the square root term

a2+x2

we may be able to trigonometric substitution to solve the integral.

Example 2.2.1

Solve

1x2dx

by substituting x=sinθ and dx=cosθdθ.

The integrand then becomes

(2.2.1)1x2=1sin2θ(2.2.2)=cos2θ(2.2.3)=cosθ

We have

(2.2.4)1x2dx=cosθcosθdθ(2.2.5)=cos2θdθ(2.2.6)=(12+12cos2θ)dθ(2.2.7)=12θ+14sin2θ+C(2.2.8)=12arcsinx+12sinθcosθ+C(2.2.9)=12arcsinx+12x1x2+C

Exercises 2.2.1
  1. 1x2x4dx
  2. 149x2dx

Two Key Formulas

From Trigonometry, we have the following two key formulas:

sec2x=1+tan2x

so

secx=1+tan2x

and

tan2x=sec2x1

so

tanx=sec2x1.

When we have integrals that involve any of the above square roots, we can use the appropriate substitution.

Example 2.2.2

(2.2.10)x31+x2dx(2.2.11)x=tanθ,dx=sec2θdθ(2.2.12)1+x2=1+tan2θ=sec2θ=secθ(2.2.13)=tan3θsecθsec2θdθ(2.2.14)=tan3θsecθdθ(2.2.15)=tan2θtanθsecθdθ(2.2.16)=(sec2θ1)secθtanθdθ(2.2.17)u=secθ,du=secθtanθdθ(2.2.18)=(u21)du(2.2.19)=u33u+C(2.2.20)=sec3θ3secθ+C(2.2.21)=(1+x2)3231+x2+C

Exercise 2.2.1
  1. x3x21dx
  2. x29+4x2dx
  3. 1x2+2xdx

Larry Green (Lake Tahoe Community College)

  • Integrated by Justin Marshall.


This page titled 2.2: Trigonometric Substitution is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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