Trigonometric substitution

Section 1.20 Trigonometric substitution

Trigonometric substitution is an example of a more general integral technique that we call inverse substitution. In turn, inverse substitution is really just an application of the substitution where our substitution function is invertible.

Procedure 1.20.1. Inverse substitution: indefinite integral.

To compute \(\displaystyle\int f(x)\, dx\) using inverse substitution, proceed as follows.

Choose a one-to-one, differentiable substitution function \(g\) with differentiable inverse and assemble the two equations

\begin{align*} x\amp =g(t)\\ dx\amp =g'(t)\, dt \end{align*}
Compute

\begin{equation*} \int f(g(t))g'(t)\, dt=F(t)+C\text{.} \end{equation*}
Conclude that

\begin{equation*} \int f(x)\, dx=F(g^{-1}(x))+C\text{.} \end{equation*}

When an explicit formula for \(g^{-1}(x)\) is not available, we attempt to compute \(F(g^{-1}(x))\) by rewriting \(F(t)\) in terms of \(x\) using the equation \(x=g(t)\text{.}\)

Procedure 1.20.2. Trigonometric substitution.

The table below indicates potentially helpful inverse substitutions for functions \(f\) of various types.

\begin{align*} f(x) \text{ contains } \sqrt{a^2-x^2}\amp \implies \begin{array}{rcl} x\amp =\amp a\sin\theta \\ dx\amp = \amp a\cos\theta\, d\theta \end{array} \hspace{5pt} \text{ where } \theta\in [-\pi/2, \pi/2]\\ \amp \\ f(x) \text{ contains } \hspace{10pt} x^2+a^2 \amp \implies \begin{array}{rcl} x\amp =\amp a\tan\theta \\ dx\amp = \amp a\sec^2\theta\, d\theta \end{array} \hspace{5pt} \text{ where } \theta\in (-\pi/2, \pi/2)\\ \amp \\ f(x) \text{ contains } \sqrt{x^2-a^2} \amp \implies \begin{array}{rcl} x\amp =\amp a\sec\theta \\ dx\amp = \amp a\sec \theta\tan\theta \, d\theta \end{array} \hspace{5pt} \text{ where } \theta\in [0,\pi/2)\cup (\pi/2, \pi]\text{.} \end{align*}

Procedure 1.20.3. Reference triangles.

When finishing an integral computation \(\int f(x)\, dx\) using an trigonometric substitution, the following types of reference triangles are useful for converting your expression back in terms of \(x\text{.}\)

For the sake of space, we only provide examples of reference triangles where the angle \(\theta\) lies in \([0,\pi/2]\text{.}\) In practice, you should provide a reference triangle diagram that reflects the interval \(\theta\) is restricted to, and take care with the sign of the various trigonometric values.

Sine substitution.

Consider the substitution \(x=a\sin \theta\text{,}\) where \(a\) is positive. A reference diagram like the following helps us see that

\begin{align*} \cos\theta \amp = \frac{\sqrt{a^2-x^2}}{a} \amp \tan\theta\amp = \frac{x}{a}\text{.} \end{align*}

Figure 1.20.4. Reference triangle for \(x=a\sin\theta\)
Tangent substitution.

Consider the substitution \(x=a\tan \theta\text{,}\) where \(a\) is positive. A reference diagram like the following allows us to deduce that

\begin{align*} \cos\theta \amp = \frac{a}{\sqrt{x^2+a^2}} \amp \sin\theta\amp =\frac{x}{\sqrt{x^2+a^2}}\text{.} \end{align*}

Figure 1.20.5. Reference triangle for \(x=a\tan\theta\)
Secant substitution.

Consider the substitution \(x=a\sec \theta\text{,}\) where \(a\) is positive. A reference diagram like the following allows us to deduce that

\begin{align} \sin \theta\amp =\frac{\sqrt{a^2-x^2}}{\abs{x}} \amp \tan \theta\amp =\begin{cases} \frac{\sqrt{x^2-a^2}}{a} \amp \text{if } \theta\in [0,\pi/2) \\ -\frac{\sqrt{x^2-a^2}}{a} \amp \text{if } \theta\in (\pi/2, \pi]. \end{cases}\tag{1.20.1} \end{align}

Figure 1.20.6. Reference triangle for \(x=a\sec\theta\)

Procedure 1.20.7. Inverse substitution: definite integral.

To compute \(\displaystyle\int_a^b f(x)\, dx\) using inverse substitution, proceed as follows.

Choose a one-to-one, differentiable substitution function \(g\) with differentiable inverse and write down the two inverse substitution equations

\begin{align*} x\amp =g(t)\\ dx\amp =g'(t)\, dt \end{align*}
Conclude that

\begin{equation*} \displaystyle\int_{x=a}^{x=b} f(x)\, dx=\int_{t=g^{-1}(a)}^{t=g^{-1}(b)}f(g(t))g'(t)\, dt\text{.} \end{equation*}

Example 1.20.8. Sine substitution.

Find an antiderivative of \(f(x)=\sqrt{1-x^2}\text{.}\)

Solution.

Using Procedure 1.20.2 we try the inverse substitution \(x=\sin \theta, dx=\cos \theta\, d\theta\) with restriction \(-\pi/2\leq \theta\leq \pi/2\text{.}\) We compute

\begin{align*} \int \sqrt{1-x^2}\, dx \amp = \int \sqrt{1-\sin^2\theta}\, \cos\theta\, d\theta\\ \amp = \int \sqrt{\cos^2\theta} \, \cos\theta\, d\ theta\\ \amp = \int \abs{\cos \theta}\cos \theta\, d\theta\\ \amp = \int \cos^2\theta\, d\theta \amp (\theta\in [-\pi/2,\pi/2]\implies \cos \theta\geq 0)\\ \amp = \frac{1}{2}\theta+\frac{1}{4}\sin(2\theta) +C\\ \amp = \frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta +C\\ \amp = \frac{1}{2}\arcsin(x)+\frac{1}{2}\sin(\arcsin(x))\cos(\arcsin x)+C\\ \amp = \frac{1}{2}\arcsin x+ \frac{1}{2}x\sqrt{1-x^2}+C\text{.} \end{align*}

Here we have used once again that \(\cos(\arcsin x)=\sqrt{1-x^2}\text{.}\) Recall that this can be derived using the following reference triangle.

Reference triangle for \(x=\sin\theta\)

Example 1.20.9. Area of circle.

Derive the area formula for a circle of radius \(r\text{.}\)

Solution.

Let \(\mathcal{C}\) be a circle of radius \(r\text{.}\) Choosing the origin to be the center of \(\mathcal{C}\text{,}\) we may assume that \(\mathcal{C}\) has equation \(x^2+y^2=r^2\text{.}\) The interior of the circle is the region \(\mathcal{R}\) lying between the graphs of the two function \(y=f(x)=+\sqrt{r^2-x^2}\) and \(y=g(x)=-\sqrt{r_2-x^2}\) from \(x=-r\) to \(x=r\text{.}\) Using Definition 1.8.3 we compute

\begin{align*} \operatorname{area}\mathcal{R} \amp = \int_{-r}^r\sqrt{r^2-x^2}-(-\sqrt{r^2-x^2}) \, dx \\ \amp = 2\int_{-r}^{r}\sqrt{r^2-x^2}\, dx\\ \amp = 2\int_{-\pi/2}^{\pi/2}\sqrt{r^2-r^2\sin^2\theta }\, r\cos \theta\, d\theta \amp (x=r\sin\theta, dx=r\cos\theta dx)\\ \amp 2r^2\int_{-\pi/2}^{\pi/2}\sqrt{\cos^2\theta}\cos\theta\, d\theta\\ \amp 2r^2\int_{-\pi/2}^{\pi/2}\abs{\cos\theta}\cos\theta\, d\theta\\ \amp = 2r^2\int_{-\pi/2}^{\pi/2}\cos^2\theta\, d\theta \amp (\theta\in [-\pi/2,\pi/2]\implies \cos\theta > 0)\\ \amp = 2r^2(\frac{1}{2}\theta+\frac{1}{2}\sin 2\theta)\Bigr\vert_{-\pi/2}^{\pi/2} \amp (\cos^2\theta=\frac{1}{2}(1+\cos 2\theta))\\ \amp = 2r^2(\pi/4-(-\pi/4))\\ \amp = \pi r^2\text{.} \end{align*}

Example 1.20.10. Tangent substitution.

Compute \(\displaystyle\int \frac{1}{x^2\sqrt{x^2+4}}\, dx\text{.}\)

Solution.

Following Procedure 1.20.2, we try the inverse substitution \(x=2\tan\theta\text{,}\) \(dx=2\sec^2\theta dx\text{,}\) \(\theta\in (-\pi/2, \pi/2)\text{:}\)

\begin{align*} \int \frac{1}{x^2\sqrt{x^2+4}}\, dx \amp = \int \frac{1}{4\tan^2 \theta\sqrt{4\tan^2\theta+4}}\cdot 2\sec^2\theta\, d\theta \\ \amp = \frac{1}{4}\int \frac{1}{\tan^2\theta\sqrt{\sec^2\theta}}\cdot \sec^2\theta\, d\theta\\ \amp = \frac{1}{4}\int \frac{1}{\tan^2\theta\abs{\sec\theta}}\cdot \sec^2\theta\, d\theta\\ \amp =\frac{1}{4}\int \frac{1}{\tan^2\theta \sec\theta}\cdot \sec^2\theta\, d\theta \amp (\theta\in (-\pi/2, \pi/2)\implies \sec\theta > 0)\\ \amp =\frac{1}{4}\int \frac{\sec\theta}{\tan^2\theta}\, d\theta\\ \amp = \frac{1}{4}\int \frac{\cos \theta}{\sin^2\theta}\, d\theta\\ \amp = -\frac{1}{4\sin\theta}+C\\ \amp = -\frac{\sqrt{x^2+4}}{4x}+C\text{.} \end{align*}

Here we deduce that \(\sin\theta=\frac{x}{\sqrt{x^2+4}}\) from the reference triangle for \(x=2\tan\theta\text{.}\)

Reference triangle for tangent substitution

Example 1.20.11. Secant substitution.

Let \(a\) be a fixed positive number. Compute \(\displaystyle \int \frac{1}{\sqrt{x^2-a^2}}\, dx\text{.}\)

Solution.

Following Procedure 1.20.2 we try the inverse substitution \(x=a\sec\theta\text{,}\) \(dx=a\sec\theta\tan\theta\text{,}\) \(\theta\in [0,\pi/2)\cup (\pi/2, \pi]\text{:}\)

\begin{align*} \int \frac{1}{\sqrt{x^2-a^2}}\, dx \amp = \int \frac{1}{\sqrt{a^2\sec^2 x-a^2}}\cdot a\sec\theta\tan\theta\, dx \\ \amp =\int \frac{1}{\sqrt{\tan^2 \theta}}\sec\theta\tan\theta\, d\theta\\ \amp = \begin{cases} \int \frac{1}{\tan\theta}\sec\theta\tan\theta\, d\theta \amp \text{if } \theta\in [0,\pi/2) \\ -\int\frac{1}{\tan\theta}\sec\theta\tan\theta\, d\theta \amp \text{if }\theta\in (\pi/2,\pi] \end{cases} \amp (\knowl{./knowl/xref/ref_tri_sec.html}{\text{Secant substitution}})\\ \amp = \begin{cases} \int \sec\theta\, d\theta \amp \text{if } \theta\in [0,\pi/2) \\ -\int\sec\theta\, d\theta \amp \text{if }\theta\in (\pi/2,\pi] \end{cases}\\ \amp = \begin{cases} \ln\abs{\sec\theta+\tan\theta}+C \amp \text{if } \theta\in [0,\pi/2) \\ -\ln\abs{\sec\theta+\tan\theta}+C \amp \text{if }\theta\in (\pi/2,\pi] \end{cases}\\ \amp =\begin{cases} \ln\abs{\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}}+C \amp \text{if } \theta\in [0,\pi/2) \\ -\ln\abs{\frac{x}{a}-\frac{\sqrt{x^2-a^2}}{a}}+C \amp \text{if }\theta\in (\pi/2,\pi] \end{cases} \amp (\knowl{./knowl/xref/ref_tri_sec.html}{\text{Secant substitution}})\\ \amp =\begin{cases} \ln\abs{\frac{x+\sqrt{x^2-a^2}}{a}}+C \amp \text{if } \theta\in [0,\pi/2) \\ -\ln\abs{\frac{x-\sqrt{x^2-a^2}}{a}}+C+C \amp \text{if }\theta\in (\pi/2,\pi] \end{cases} \\ \amp =\begin{cases} \ln\abs{x+\sqrt{x^2-a^2}}-\ln a+C \amp \text{if } \theta\in [0,\pi/2) \\ \ln\abs{\frac{1}{x-\sqrt{x^2-a^2}}}-\ln a+C \amp \text{if }\theta\in (\pi/2,\pi] \end{cases} \amp (\text{algebra}) \\ \amp = \ln\abs{x+\sqrt{x^2-a^2}}-\ln a+C \amp \left(\frac{1}{x-\sqrt{x^2-a^2}}=x+\sqrt{x^2-a^2}\right)\\ \amp = \ln\abs{x+\sqrt{x^2-a^2}}+D\text{.} \end{align*}

Example 1.20.12. Secant substitution: definite.

Compute \(\displaystyle\int_{-\sqrt{2}}^{-2/\sqrt{3}}\frac{\sqrt{x^2-1}}{x}\, dx\text{.}\)

Solution.

We use the inverse substitution \(x=\sec\theta, dx=\sec\theta\tan\theta\text{.}\) Solve \(-1/\sqrt{2}=\sec\theta\) and \(-2/\sqrt{3}=\sec\theta\) for \(\theta\in [0,\pi/2)\cup (\pi/2,\pi]\text{,}\) we see that the new limits of integration are \(\theta=3\pi/4\) and \(\theta=5\pi/6\text{.}\) Thus we have

\begin{align*} \int_{-\sqrt{2}}^{-2/\sqrt{3}}\frac{\sqrt{ \theta^2-1}}{ \theta}\, d \theta \amp = \int_{3\pi/4}^{5\pi/6}\frac{\sqrt{\sec^2 \theta-1}}{\sec \theta}\cdot \sec \theta\tan \theta\, d\theta \\ \amp =\int_{3\pi/4}^{5\pi/6}\frac{\abs{\tan \theta}}{\sec \theta}\cdot \sec \theta\tan \theta\, d\theta \\ \amp = \int_{3\pi/4}^{5\pi/6}-\tan^2 \theta\, d\theta \amp (\theta\in (\pi/2, \pi]\implies \tan\theta< 0)\\ \amp = \int_{3\pi/4}^{5\pi/6}1-\sec^2 \theta\, d\theta\\ \amp = \theta-\tan\theta\Bigr\vert_{3\pi/4}^{5\pi/6}\\ \amp = \frac{\pi}{12}+\frac{1}{\sqrt{3}}-1\text{.} \end{align*}

Example 1.20.13. Secant substitution: indefinite.

Find an antiderivative \(F\) of the function \(f(x)=\frac{\sqrt{x^2-1}}{x}\, dx\) on the domain \((-\infty,-1]\cup [1,\infty)\text{.}\) The only inverse trigonometric functions allowed to appear in your formula for \(F\) are \(\arcsin x\) and \(\arctan x\text{.}\)

Solution.

To be assigned on written homework.

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