8.4.2 Trigonometric Substitutions

Now we take advantage of our knowledge of inverse trigonometric functions to use substitution in a slightly different way as an aid in evaluating integrals. The formulas from Chapter 5 for derivatives of the inverse sine and inverse tangent will play a role here, especially in checking our antiderivative calculations:

d d ⁢ x ⁢ sin -1 ( x ) = 1 1 - x 2,

d d ⁢ x ⁢ tan -1 ( x ) = 1 1 + x 2.

Example 1

Evaluate

Solution   We would like to make a substitution that would turn \(25-x^2\) into the square of something. An identity that looks useful is

For our purposes, it would be more useful in the form

or

Thus we could substitute \(x=5\,\sin(\theta)\). The new variable \(\theta\) we are introducing into the problem is defined by the inverse relationship, \(\displaystyle\theta = \sin^{-1}\left(\frac{x}{5}\right)\). From the description \(x=5\, \sin(\theta)\), we obtain \(dx = 5\,\cos(\theta)\,d \theta\). When we make the substitutions for \(x\) and \(dx\), we find

Now we would like to say that \(\sqrt{\cos^2(\theta)}=\cos(\theta)\). However, that is not always true - it holds only if we know that \(\theta\) is restricted to a range where \(\cos(\theta)\) is nonnegative, such as the interval between \(-\,\pi/2\) and \(\pi/2\). But the \(\theta\)'s we are considering are already restricted to this interval because they are values of the arcsine function. Thus, in this setting, it is true that \(\sqrt{\cos^2(\theta)}=\cos(\theta)\).

We are left with the simple indefinite integral

Replacing \(\theta\) with \(\displaystyle\sin^{-1}\left(\frac{x}{5}\right)\), we find

Checkpoint 2

Checkpoint 3

Example 2

Evaluate

Solution   When we make the substitution \(x=5\,\sin(\theta)\), we obtain

Moreover, when \(x=0\), \(\displaystyle\theta=\sin^{-1}\left(\frac{0}{5}\right)=0\), and when \(\displaystyle x=\frac{5}{\sqrt{2}}\), \(\displaystyle\theta=\sin^{-1}\frac{1}{\sqrt{2}}=\frac{\pi}{4}\). Thus we may replace the original integral by

To evaluate this replacement integral, we need an antiderivative for \(\cos^2(\theta)\). The following identities are helpful in integrating squares of sines and squares of cosines:

Using the latter identity, we find

Thus the integral we want may be evaluated by using the Fundamental Theorem of Calculus:

Checkpoint 4