Archive

Integrals involving trig substitutions

No Comments

 

When is it appropriate to solve an integral with a trig substitution? First of all, keep in mind that a trig substitution doesn’t always work. Even when it does work, you are often left with an integral that will require other techniques such as a u substitution or integration by parts. But if you are willing to put in a little effort (and you know your trig identities), trig substitutions allow you to find the antiderivatives of some rather complicated functions.

There are three conditions that you look for—each a radical term of a particular form in the integrand. Each condition is associated with a different substitution. After you make the substitution, you simplify the integrand and go from there.

Term

Substitution

Radical becomes…

\sqrt{a^2-x^2} \text{Let } x=a \sin \theta; \, dx=a \cos \theta \,d \theta a \cos \theta
\sqrt{a^2+x^2} \text{Let } x=a \tan \theta; \, dx=a \sec^2 \theta \,d \theta a \sec \theta
\sqrt{x^2-a^2} \text{Let } x=a \sec \theta; \, dx=a \sec \theta \tan \theta \,d \theta a \tan \theta

 

Before we look at some example integrals, let’s see why the first radical term above simplifies to a \cos \theta. It’s pretty straightforward if you know your trig identities:

\sqrt{a^2-x^2}=\sqrt{a^2-(a \sin \theta)^2}= \sqrt{a^2-a^2 \sin^2 \theta}= \sqrt{a^2(1- \sin^2 \theta)} =

 …….. \sqrt{a^2 \cos^2 \theta} = a \cos \theta

Example. Integrate the following:

\text{a.} \displaystyle \int \frac{dx}{\sqrt{1-x^2}} \qquad \text{b.} \int \frac{x^3}{8 \sqrt{4+x^2}}dx \qquad \text{c.} \int x \sqrt{x^2-4} \; dx

Solutions:

  1. (Does this integral look familiar?) Here, a=1, so use x= \sin \theta. Using the first line of the table above:
    …..
    \displaystyle \int \frac{dx}{\sqrt{1-x^2}}= \int \frac{\cos \theta \, d \theta}{\cos \theta}= \int d \theta = \theta +C
    …..
    But since x= \sin \theta, \theta=\sin^{-1} x.
    …..
    \therefore \displaystyle \int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1} x+C
  2. Here, a=2 and we use line 2 from the table above (x=2 \tan \theta). Note that x^3=8 \tan^3 \theta. Upon substitution,
    …..
    \displaystyle \int \frac{x^3}{8 \sqrt{4+x^2}}\, dx= \int \frac{8 \tan^3 \theta}{8(2 \sec \theta)}2 \sec^2 \theta \, d \theta = \int \tan^3 \theta \sec \theta \, d \theta
    …..
    Hmm. This is going to take a little bit of extra work… Time to pull out some trig identities:
    …..
    \int \tan^3 \theta \sec \theta \, d \theta = \int \tan^2 \theta \tan \theta \sec \theta \, d \theta = \int (\sec^2 \theta -1) \tan \theta \sec \theta \, d \theta
    …..
    Now a u substitution, letting u= \sec \theta:
    …..
    \displaystyle \int (u^2-1) \, du= \frac{1}{3} u^3-u+C= \frac{1}{3} \sec^3 \theta - \sec \theta +C
    …..
    How do we get our answer back in terms of x? Draw a triangle that shows how x and \tan \theta are related, then use the Pythagorean theorem to find an expression for \sec \theta. In the triangle below,
    …..
    x=2 \tan \theta \therefore \sec \theta = \dfrac{\sqrt{4+x^2}}{2}
    Substitute into the integral above to get:
    \displaystyle \int \frac{x^3}{8 \sqrt{4+x^2}} dx=\frac{1}{3} \sec^3 \theta - \sec \theta +C= \frac{(4+x^2)^{3/2}}{24}- \frac{(4+x^2)^{1/2}}{2}+C

    This can be simplified further by factoring:
    …..
    \displaystyle \int \frac{x^3}{8 \sqrt{4+x^2}}dx= \frac{(4+x^2)^{3/2}}{24}- \frac{(4+x^2)^{1/2}}{2}+C= \frac{1}{24}(x^2-8) \sqrt{4+x^2}
    …..
    Whew!

  3. This is a trick question. Even though it fits the condition given in the table (and you could integrate with a trig substitution if you wanted), it’s easier to do this one with a u substitution: u=x^2-4 and du=2x \; dx:
    …..
    \displaystyle \int x \sqrt{x^2-4} \, dx= \frac{1}{2} \int \sqrt{u} \, du= \frac{1}{3}u^{3/2}+C= \frac{1}{3}(x^2-4)^{3/2}+C

The lesson here is to look for u substitutions before you look for trig substitutions.

Blue Taste Theme created by Jabox