7. Trigonometric Substitutions
a3. Substitutions for \(x^2-1\) (or \(1-x^2\)) - Secant Substitutions
Recall the derivative and integral formulas for \(\mathrm{arcsec}\,x\): \[ \dfrac{d}{dx}\mathrm{arcsec}\,x=\dfrac{1}{|x|\sqrt{x^2-1}} \qquad \qquad \qquad \int \dfrac{1}{|x|\sqrt{x^2-1}}\,dx=\mathrm{arcsec}\,x+C \] and the second Pythagorean Trig Identity written as: \[ \sec^2\theta-1=\tan^2\theta \]
These formulas motivate the following integration technique:
If the integrand involves the quantity \(x^2-1\) (or \(1-x^2\)) and
you know that \(|x| \ge 1\), then it may be useful to make the
substitution \(x=\sec\theta\). Then \(dx=\sec\theta\tan\theta\,d\theta\)
and \(x^2-1=\sec^2\theta-1=\tan^2\theta\). So it may be possible to
re-express the integrand in terms of \(\sec\theta\) and \(\tan\theta\) only.
Why do you need to know \(|x| \ge 1\)?
Since \(|\sec\theta| \ge 1\), if \(|x| \lt 1\) then we cannot set \(x=\sec\theta\).
Graph of \(\sec\theta\).
How do you know if \(|x| \ge 1\)?
You look at any quantities inside square roots such as \(\sqrt{x^2-1}\) or at the limits on the integral.
Here are some examples:
Compute \(\displaystyle \int \dfrac{(x^2-1)^{3/2}}{x}\,dx\).
Since \(x^2-1\) is raised to the \(\dfrac{3}{2}\) power, we must have \(x^2-1 \ge 0\). So \(x^2 \ge 1\) or \(|x| \ge 1\). So we substitute \(x=\sec\theta\). Then \(dx=\sec\theta\tan\theta\,d\theta\) and: \[\begin{aligned} \int \dfrac{(x^2-1)^{3/2}}{x}\,dx &=\int \dfrac{(\sec^2\theta-1)^{3/2}}{\sec\theta} \sec\theta\tan\theta\,d\theta \\ &=\int \dfrac{\tan^3\theta}{\sec\theta} \sec\theta\tan\theta\,d\theta =\int \tan^4\theta\,d\theta \end{aligned}\] We now use the identity \(\tan^2\theta=\sec^2\theta-1\) twice to write: \[\begin{aligned} \tan^4\theta &=\tan^2\theta(\sec^2\theta-1) =\tan^2\theta\sec^2\theta-\tan^2\theta \\ &=\tan^2\theta\sec^2\theta-\sec^2\theta+1 \end{aligned}\] So: \[\begin{aligned} \int \dfrac{(x^2-1)^{3/2}}{x}\,dx &=\int \tan^2\theta\sec^2\theta-\sec^2\theta+1\,d\theta \\ &=\dfrac{\tan^3\theta}{3}-\tan\theta+\theta+C \end{aligned}\] Don't forget to substitute for the differential! Don't forget to substitute back! We now substitute back. We already know that \(\sec\theta=x\) from the definition of the substitution. So: \[ \theta=\mathrm{arcsec}\,x \] We need a formula for \(\tan\theta\) in terms of \(x\). There are two ways to find this formula:
We can use the Pythagorean identity \(\tan^2\theta=\sec^2\theta-1\). So: \[ \tan\theta=\sqrt{\sec^2\theta-1}=\sqrt{x^2-1} \]
We can draw a right triangle with an angle \(\theta\) whose hypotenuse is \(x\) and whose adjacent side is \(1\). These sides are chosen so that \(\sec\theta=\dfrac{x}{1}=x\). Then the opposite side is \(\sqrt{x^2-1}\) and so: \[ \tan\theta=\sqrt{x^2-1} \]
In either case the original integral is: \[\begin{aligned} \int \dfrac{(x^2-1)^{3/2}}{x}\,dx &=\dfrac{\tan^3\theta}{3}-\tan\theta+\theta+C \\ &=\dfrac{(x^2-1)^{3/2}}{3}-\sqrt{x^2-1}+\mathrm{arcsec}\,x+C \end{aligned}\]
We check by differentiating. If \(f=\dfrac{(x^2-1)^{3/2}}{3}-\sqrt{x^2-1}+\mathrm{arcsec}\,x\), then \[\begin{aligned} f'(x) &=x(x^2-1)^{1/2}-\dfrac{x}{\sqrt{x^2-1}}+\dfrac{1}{x\sqrt{x^2-1}} \\ &=\dfrac{x^2(x^2-1)-x^2+1}{x\sqrt{x^2-1}} =\dfrac{x^4-2x^2+1}{x\sqrt{x^2-1}} \\ &=\dfrac{(x^2-1)^2}{x\sqrt{x^2-1}} =\dfrac{(x^2-1)^{3/2}}{x} \end{aligned}\] which is the integrand we started with.
Compute \(\displaystyle \int_{\sqrt{2}}^2 \dfrac{1}{1-x^2}\,dx\).
From the limits, we see that \(\sqrt{2} \le x \le 2\) and hence \(|x| \gt 1\). So we substitute \(x=\sec\theta\). Then \(dx=\sec\theta\tan\theta\,d\theta\). Rather than substituting back, we change limits. So: \[\begin{aligned} \int_{\sqrt{2}}^2 &\dfrac{1}{1-x^2}\,dx =\int_{\pi/4}^{\pi/3} \dfrac{1}{1-\sec^2\theta}\sec\theta\tan\theta\,d\theta \\ &=\int_{\pi/4}^{\pi/3} \dfrac{1}{-\tan^2\theta}\sec\theta\tan\theta\,d\theta =-\int_{\pi/4}^{\pi/3} \dfrac{\sec\theta}{\tan\theta}\,d\theta \\ &=-\int_{\pi/4}^{\pi/3} \dfrac{1}{\sin\theta}\,d\theta =-\int_{\pi/4}^{\pi/3} \csc\theta\,d\theta \\ &=-\left[\dfrac{}{}-\ln|\csc\theta+\cot\theta|\right]_{\pi/4}^{\pi/3} \\ \end{aligned}\]
We check by differentiating. If \(f=-\ln|\csc\theta+\cot\theta|\), then \[\begin{aligned} f'(x) &=-\,\dfrac{-\csc\theta\cot\theta-\csc^2\theta}{\csc\theta+\cot\theta} \\ &=\dfrac{\csc\theta(\cot\theta+\csc\theta)}{\csc\theta+\cot\theta} =\csc\theta \end{aligned}\] This is a formula to remember like the one for \(\displaystyle \int \sec\theta\,d\theta\).
Finally, we evaluate at the limits: \[\begin{aligned} \int_{\sqrt{2}}^2 \dfrac{1}{1-x^2}\,dx &=\ln\left|\csc\dfrac{\pi}{3}+\cot\dfrac{\pi}{3}\right| -\ln\left|\csc\dfrac{\pi}{4}+\cot\dfrac{\pi}{4}\right| \\ &=\ln\left|\dfrac{2}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}\right| -\ln|\sqrt{2}+1| \\ &=\ln\sqrt{3}-\ln(\sqrt{2}+1) \end{aligned}\]
Now it's your turn:
Compute \(\displaystyle \int \dfrac{\sqrt{x^2-1}}{x}\,dx\).
Let \(x=\sec\theta\).
\(\displaystyle \int \dfrac{\sqrt{x^2-1}}{x}\,dx =\sqrt{x^2-1}-\mathrm{arcsec}\,x+C\)
We substitute \(x=\sec\theta\). Then \(dx=\sec\theta\tan\theta\,d\theta\) and: \[\begin{aligned} \int \dfrac{\sqrt{x^2-1}}{x}\,dx &=\int \dfrac{\sqrt{\sec^2\theta-1}}{\sec\theta}\sec\theta\tan\theta\,d\theta =\int \tan^2\theta\,d\theta \\ &=\int \sec^2\theta-1\,d\theta =\tan\theta-\theta +C\\ \end{aligned}\] To substitute back, we have \(\sec\theta=x\) and so \(\theta=\mathrm{arcsec}\,x\). From either the Pythagorean identity or the triangle plot, we have: \[ \tan\theta=\sqrt{\sec^2\theta-1}=\sqrt{x^2-1} \] So: \[ \int \dfrac{\sqrt{x^2-1}}{x}\,dx =\sqrt{x^2-1}-\mathrm{arcsec}\,x+C \]
We check by differentiating. If \(f(x)=\sqrt{x^2-1}-\mathrm{arcsec}\,x\), then \[\begin{aligned} f'(x) &=\dfrac{1}{2}\dfrac{2x}{\sqrt{x^2-1}}-\dfrac{1}{x\sqrt{x^2-1}} \\ &=\dfrac{x^2-1}{x\sqrt{x^2-1}} \\ &=\dfrac{\sqrt{x^2-1}}{x} \end{aligned}\] which is the integrand we started with.
Compute \(\displaystyle \int \dfrac{x}{(1-x^2)^2}\,dx\) assuming \(|x|>1\).
Let \(x=\sec\theta\). Or use the ordinary substitution \(u=1-x^2\).
\(\displaystyle \int \dfrac{x}{(1-x^2)^2}\,dx =\dfrac{1}{2(1-x^2)}+C\)
We substitute \(x=\sec\theta\) and \(dx=\sec\theta\tan\theta\,d\theta\): \[\begin{aligned} \int \dfrac{x}{(1-x^2)^2}\,dx &=\int \dfrac{\sec\theta}{(1-\sec^2\theta)^2}\sec\theta\tan\theta\,d\theta \\ &=\int \dfrac{\sec^2\theta}{\tan^3\theta}\,d\theta \end{aligned}\] Now let \(u=\tan\theta\). Then \(du=\sec^2\theta\,d\theta\). So: \[ \int \dfrac{x}{(1-x^2)^2}\,dx =\int \dfrac{1}{u^3}\,du =\dfrac{1}{-2u^2}+C \] Finally we substitute back: \[\begin{aligned} \int \dfrac{x}{(1-x^2)^2}\,dx &=-\,\dfrac{1}{2\tan^2\theta}+C =-\,\dfrac{1}{2(\sec^2\theta-1)}+C \\ &=\dfrac{1}{2(1-x^2)}+C \end{aligned}\] This solution was unnecessarily complicated. Always look for an ordinary substitution first:
We substitute \(u=1-x^2\). Then \(du=-2x\,dx\) and \(-\,\dfrac{1}{2}\,du=x\,dx\). So, \[\begin{aligned} \int \dfrac{x}{(1-x^2)^2}\,dx &=-\,\dfrac{1}{2}\int \dfrac{1}{u^2}\,du =\dfrac{1}{2u}+C \\ &=\dfrac{1}{2(1-x^2)}+C \end{aligned}\]
We check by differentiating. If \(f(x)=\dfrac{1}{2(1-x^2)}\), then \[\begin{aligned} f'(x) &=-\,\dfrac{1}{2}\dfrac{-2x}{(1-x^2)^2} \\ &=\dfrac{x}{(1-x^2)^2} \end{aligned}\] which is the integrand we started with.
It is interesting that the integrand and the answer to this last exercise (which used a \(\sec\) substitution) are the same as that for the last exercise on the previous page (which used a \(\sin\) substitution). However, it is not surprising since they can both be done with the same ordinary substitution. On the other hand, if you just ask for the antiderivative of \(\dfrac{x}{(1-x^2)^2}\), then the answer is \(\dfrac{1}{2(1-x^2)}+C\), but the constant of integration \(C\) might be different in each of the three intervals \((-\infty,-1)\), \((-1,1)\) and\((1,\infty)\).
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