Trigonometric Substitution

Calculus 2

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Overview

Some integrals solve much more easily when we perform a trigonometric substitution. In particular, trigonometric substitution is great for getting rid of radicals in our integrand (given that our tried and true u-substitution isn't an option).

There are three substitutions we make use of:

x=absin(θ)    x=absec(θ)    x=abtan(θ)x = \frac{a}{b}\sin(\theta) \ \ \ \ x=\frac{a}{b}\sec(\theta) \ \ \ \ x=\frac{a}{b}\tan(\theta)

The goal when making one of these substitutions is to obtain a Pythagorean identity for which the radical will reduce (although the radical is not necessary, as we will see later):

Things to keep in mind:

  • We are making a substitution for xx, so also replace dxdx
  • Always put our answer back into terms of xx
  • When simplifying, try rewriting everything in terms of sines and cosines
  • Don't forget to consider the angle restrictions (more on this when we discuss definite integrals)

Indefinite Integrals

Example 1

Compute 4x2x2dx.\int \frac{\sqrt{4-x^2}}{x^2} \, dx.

View Answer

4x2xarcsin(x2)+C.-\frac{\sqrt{4-x^2}}{x} - \arcsin\left(\frac{x}{2}\right) + C.

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Example 2

Compute 1xx29dx.\int \frac{1}{x\sqrt{x^2-9}} \, dx.

View Answer

13sec1(x3)+C.\frac{1}{3} \sec^{-1}\left(\frac{x}{3}\right) + C.

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Example 3

Compute 1x2x2+25dx.\int \frac{1}{x^2\sqrt{x^2+25}} \, dx.

View Answer

x2+2525x+C.-\frac{\sqrt{x^2+25}}{25x} + C.

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Definite Integrals

Example 4

Compute 12x21xdx.\int_1^2 \frac{\sqrt{x^2-1}}{x} \, dx.

View Answer

3π3.\sqrt{3} - \frac{\pi}{3}.

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Without Radicals

Note that trigonometric substitution can also be used in cases where there isn’t a radical in the integrand.

Example 5

Compute 1(x2+1)2dx.\int \frac{1}{(x^2+1)^2} \, dx.

View Answer

12arctanx+x2(x2+1)+C.\frac{1}{2}\arctan x + \frac{x}{2(x^2+1)} + C.

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Challenge 1

Set up the integral for the arc length of y=lnxy = \ln x from x=1x=1 to x=3x=3.

Then, state the trigonometric substitution that would be used to solve the problem.

View Answer

131+1x2dx=13x2+1xdx,let x=tanθ.\int_1^3 \sqrt{1+\frac{1}{x^2}} \, dx = \int_1^3 \frac{\sqrt{x^2+1}}{x} \, dx, \quad \text{let } x = \tan{\theta}.

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