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