This is a new method of substitution, and at first, it can be difficult to get used to. The biggest difference from the substitution methods we've used so far (u-substitution and integration by parts) is that rather than defining a dummy variable *explicitly* in terms of \(x\), like \(u= \) "some expression of \(x\)," we'll define the dummy variable (\(\theta\) in this case) *implicitly* by writing \(x= \) "some expression of \(\theta\)."

Just like before, we'll do this so that the integral becomes something we can handle (usually by means of trig integration, as outlined in the previous section), and we'll have to back-substitute in order to have an answer in terms of \(x\).

Let's illustrate this process with an example.

Evaluate \(\displaystyle\int \dfrac{1}{\sqrt{1-x^2}} \ dx\).

Let \(x=\sin \theta\). Therefore, \(dx = \cos \theta \ d\theta\). Notice why this helps us: when we substitute this in, we can use the Pythagorean Identity to simplify the integral: \[\int \dfrac{1}{\sqrt{1-x^2}} \ dx = \int \dfrac{1}{\sqrt{1-\sin^2 \theta}}\ \cos \theta \ d\theta = \int \dfrac{1}{\sqrt{\cos^2 \theta}}\ \cos \theta \ d\theta = \int \dfrac{1}{\cos \theta}\ \cos \theta \ d\theta = \int d\theta = \theta + C\]

To rewrite this answer in terms of \(x\), notice that if \(x=\sin \theta\), \(\theta = \sin^{-1} x\): \[\int \dfrac{1}{\sqrt{1-x^2}} \ dx = \int d\theta = \theta + C = \ans{\sin^{-1}x + C}\]

There are three forms that we can handle with trig substitution. The key is to match each form to the appropriate trig identity (all based on the Pythagorean Identity).

- Form 1: \(\sqrt{a^2-x^2}\) for some constant \(a\). This looks like 1-something
^{2}, which reminds us of \(\cos^2 \theta = 1-\sin^2 \theta\). Therefore, let \[x = a \sin \theta \longrightarrow dx = a \cos \theta \ d\theta.\] Notice that \[\sqrt{a^2-x^2} = \sqrt{a^2-a^2\sin^2 \theta} = \sqrt{a^2(1-\sin^2 \theta)} = \sqrt{a^2\cos^2 \theta} = a\cos \theta\] - Form 2: \(\sqrt{a^2+x^2}\) for some constant \(a\). This looks like 1+something
^{2}, which reminds us of \(1+\tan^2 \theta = sec^2 \theta\). Therefore, let \[x = a \tan \theta \longrightarrow dx = a \sec^2 \theta \ d\theta.\] Notice that \[\sqrt{a^2+x^2} = \sqrt{a^2+a^2\tan^2 \theta} = \sqrt{a^2(1+\tan^2 \theta)} = \sqrt{a^2\sec^2 \theta} = a\sec \theta\] - Form 1: \(\sqrt{x^2-a^2}\) for some constant \(a\). This looks like something
^{2}-1, which reminds us of \(\tan^2 \theta = \sec^2 \theta - 1\). Therefore, let \[x = a \sec \theta \longrightarrow dx = a \sec \theta \ \tan \theta \ d\theta.\] Notice that \[\sqrt{x^2-a^2} = \sqrt{a^2\sec^2 \theta-a^2} = \sqrt{a^2(\sec^2 \theta - 1)} = \sqrt{a^2\tan^2 \theta} = a\tan \theta\]

Evaluate \(\displaystyle\int \sqrt{4-x^2} \ dx\).

Following the guidelines above, let \(x=2 \sin \theta\), so \(dx = 2\cos \theta \ d\theta\). Making the substitution: \[\begin{align} &\int \sqrt{4-x^2} \ dx = \int \sqrt{4-4\sin^2 \theta} \cdot 2\cos \theta \ d\theta\\ &= \int \sqrt{4\cos^2 \theta} \cdot 2\cos \theta \ d\theta = \int 2\cos\theta \cdot 2\cos\theta \ d\theta\\ &= \int 4\cos^2 \theta \ d\theta \end{align}\]

To evaluate this integral, go back to what we did in the trig integrals section: use a half-angle identity: \[\int 4\cos^2 \theta \ d\theta = \int 2\big(1+\cos(2\theta)\big) \ d\theta = \int 2 + 2\cos(2\theta) \ d\theta = 2\theta + \sin(2\theta) + C\]

To finish this off, we need another (heretofore unmentioned) identity: \(\sin (2\theta) = 2\sin \theta \ \cos \theta\). So \[\int \sqrt{4-x^2} \ dx = 2\theta + \sin(2\theta) + C = 2\theta + 2\sin\theta \ \cos\theta + C\]

Finally, we need to figure out what \(\sin\theta\) and \(\cos\theta\) equal in terms of \(x\). There are two common ways to do this: the first is simply to notice that \[x=2\sin\theta \longrightarrow \sin \theta = \dfrac{x}{2}\] and \[\sqrt{4-x^2} = 2\cos \theta \longrightarrow \cos \theta = \dfrac{\sqrt{4-x^2}}{2}.\] The other way is to draw a right triangle, noting that \(\sin \theta = \dfrac{x}{2}\), so we can label the side opposite to \(\theta\) as \(x\), and we can label the hypotenuse as 2. The Pythagorean Theorem lets us fill in the third side, and once we have that, we can figure out what \(\cos \theta\) is (or any other trig quantity).

This triangle gives the same information as above; namely, that \(\sin \theta = x/2\) and \(\cos \theta = \sqrt{4-x^2}/2\).

Either way we get to that point, we get the following answer: \[\int \sqrt{4-x^2} \ dx = 2\theta + 2\sin\theta \ \cos\theta + C = \ans{2\sin^{-1}\left(\dfrac{x}{2}\right) + \dfrac{1}{2}x\sqrt{4-x^2}+C}\]

Evaluate \(\displaystyle\int x^3\sqrt{9-x^2} \ dx.\)

Let \(x=3\sin \theta\), so \(dx = 3\cos \theta \ d\theta\). Also, note that \(\sqrt{9-x^2} = 3\cos \theta\). Making the substitution: \[\int x^3\sqrt{9-x^2} \ dx = \int (3\sin \theta)^3 \cdot 3\cos \theta \cdot 3\cos \theta \ d\theta = 243\int \sin^3 \theta \ \cos^2 \theta \ d\theta\]

To evaluate this integral, split off \(\sin \theta\) and write everything else in terms of \(\cos \theta\): \[243\int \sin^2 \theta \ \cos^2 \theta \ \sin \theta \ d\theta = 243\int (1-\cos^2 \theta)\cos^2 \theta \ \sin \theta \ d\theta.\]

Now, letting \(u=\cos \theta\) and \(du = -\sin\theta\), \[\begin{align} &243\int (1-\cos^2 \theta)\cos^2 \theta \ \sin \theta \ d\theta\\ &= -243\int (1-u^2)u^2 \ du\\ &= -243\int u^2-u^4 \ du\\ &= -\dfrac{243}{3}u^3 + \dfrac{243}{5}u^5 + C. \end{align}\] We have to first substitute back into terms of \(\theta\), and then back to \(x\): \[-\dfrac{243}{3}u^3 + \dfrac{243}{5}u^5 + C = -\dfrac{243}{5}\cos^3 \theta + \dfrac{243}{5}\cos^5 \theta + C\] (remember, \(\cos \theta = \sqrt{9-x^2}/3\), which we can see by either looking above at the work we've already done, or by drawing a triangle like we did in the previous example) \[\begin{align} &-\dfrac{243}{5}\cos^3 \theta + \dfrac{243}{5}\cos^5 \theta + C\\ &= -\dfrac{243}{3}\left(\dfrac{\sqrt{9-x^2}}{3}\right)^3 + \dfrac{243}{5}\left(\dfrac{\sqrt{9-x^2}}{3}\right)^5 + C\\ &= \ans{-3(\sqrt{9-x^2})^3 + \dfrac{1}{5}(\sqrt{9-x^2})^5 + C} \end{align}\]

\(\displaystyle\int \dfrac{x^3}{\sqrt{16-x^2}} \ dx\)

\(-16\sqrt{16-x^2}+\dfrac{1}{3}\big(\sqrt{16-x^2}\big)^3 + C\)

Evaluate \(\displaystyle\int \dfrac{x^3}{\sqrt{x^2+9}} \ dx.\)

This one has the second form, so we'll let \(x=3\tan \theta\), which means that \(dx = 3\sec^2 \theta \ d\theta\) and \(\sqrt{x^2+9} = 3\sec \theta\). Making the substitution, \[\int \dfrac{x^3}{\sqrt{x^2+9}} \ dx = \int \dfrac{27\tan^3 \theta}{3\sec\theta} \cdot 3\sec^2 \theta \ d\theta = 27 \int \tan^3\theta \ \sec\theta \ d\theta.\]

To evaluate this integral, split off \(\sec\theta \tan\theta\) and rewrite everything else in terms of \(\sec\theta\): \[27 \int \tan^3\theta \ \sec\theta \ d\theta = 27\int \tan^2\theta \ \sec\theta \tan\theta \ d\theta = 27\int (1-\sec^2\theta) \ \sec\theta \tan\theta \ d\theta.\] Now, letting \(u=\sec \theta\) and \(du = \sec\theta \tan\theta \ d\theta\), \[27\int (1-\sec^2\theta) \ \sec\theta \tan\theta \ d\theta = 27\int 1-u^2 \ d\theta = 27u-\dfrac{27}{3}u^3 + C.\]

When we substitute back into \(\theta\), we'll have \(\sec\theta\) in the answer. Since we found earlier that \(\sqrt{x^2+9} = 3\sec\theta\), \(\sec\theta = \sqrt{x^2+9}/3\). We could also have noted this by drawing a right triangle (you should try doing this to make sure that you can.

\[\begin{align} 27u-\dfrac{27}{3}u^3 + C &= 27\sec\theta - 9\sec^3 \theta + C\\ &= 27\left(\dfrac{\sqrt{x^2+9}}{3}\right) - 9\left(\dfrac{\sqrt{x^2+9}}{3}\right)^3 + C\\ &= \ans{9\sqrt{x^2+9} - \dfrac{1}{3}(\sqrt{x^2+9})^3 + C} \end{align}\]\(\displaystyle\int \dfrac{x^3}{\sqrt{x^2+100}} \ dx\)

\(100\sqrt{x^2+100} - \dfrac{1}{3}\big(\sqrt{x^2+100}\big)^3 + C\)

Evaluate \(\displaystyle\int \dfrac{\sqrt{x^2-1}}{x} \ dx.\)

This looks like the third form, so let \(x=\sec\theta\), which means that \(dx = \sec\theta \tan\theta \ d\theta\) and \(\sqrt{x^2-1}=\tan\theta\). Then \[\int \dfrac{\sqrt{x^2-1}}{x} \ dx = \int \dfrac{\tan\theta}{\sec\theta} \ \sec\theta \tan\theta \ d\theta = \int \tan^2 \theta \ d\theta.\]

To integrate this, rewrite \(\tan^2\theta\) as \(\sec^2\theta - 1\), then note that the antiderivative of \(\sec^2\theta\) is \(\tan\theta\). \[\int \tan^2 \theta \ d\theta = \int \sec^2\theta - 1 \ d\theta = \tan\theta - \theta + C.\]

Finally, to get the answer in terms of \(x\), note that \(\tan\theta = \sqrt{x^2-1}\) and \(\theta = \sec^{-1}x\): \[\int \dfrac{\sqrt{x^2-1}}{x} \ dx = \ans{\sqrt{x^2-1} - \sec^{-1}x + C}\]

\(\displaystyle\int x\sqrt{x^2+4} \ dx\)

\(\dfrac{1}{3} \big(\sqrt{x^2+4}\big)^3 + C\)