Section 2.3 Trigonometric Substitutions
ΒΆExample 2.26. Sine Substitution.
Evaluate β«β1βx2dx.
Let \(x=\sin u\) so \(dx=\cos u\,du\text{.}\) Then
We would like to replace \(\ds \sqrt{\cos^2 u}\) by \(\cos u\text{,}\) but this is valid only if \(\cos u\) is positive, since \(\ds \sqrt{\cos^2 u}\) is positive. Consider again the substitution \(x=\sin u\text{.}\) We could just as well think of this as \(u=\arcsin x\text{.}\) If we do, then by the definition of the arcsine, \(-\pi/2\le u\le\pi/2\text{,}\) so \(\cos u\ge0\) and so we are allowed to continue and perform the simplification:
This is a perfectly good answer, though the term \(\sin(2\arcsin x)\) is a bit unpleasant. It is possible to simplify this. Using the identity \(\sin 2x=2\sin x\cos x\text{,}\) we can write
Then the full antiderivative is
-
This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity sin2x+cos2x=1 in one of three forms:
cos2x=1βsin2xsec2x=1+tan2xtan2x=sec2xβ1.If your function contains 1βx2, as in the example above, try x=sinu; if it contains 1+x2 try x=tanu; and if it contains x2β1, try x=secu. Sometimes you will need to try something a bit different to handle constants other than inverse substitution, which is described next.
In a traditional substitution we let u=u(x), i.e., our new variable is defined in terms of x. In an inverse substitution we let x=g(u), i.e., we assume x can be written in terms of u. We cannot do this arbitrarily since we do NOT get to βchooseβ x. For example, an inverse substitution of x=1 will give an obviously wrong answer. However, when x=g(u) is an invertible function, then we are really doing a u-substitution with u=gβ1(x). Now the Substitution Rule applies.
Sometimes with inverse substitutions involving trig functions we use ΞΈ instead of u. Thus, we would take x=sinΞΈ instead of x=sinu.
We would like our inverse substitution x=g(u) to be a one-to-one function, and x=sinu is not one-to-one. In the next few paragraphs, we discuss how we can overcome this issue by using the restricted trigonometric functions.
For trig functions containing ΞΈ, use a triangle to convert to x's.
For ΞΈ by itself, use the inverse trig function.
Guideline for Trigonometric Substitution.
Suppose we have an integral with any of the following expressions, then use the substitution, differential, identity and inverse of substitution listed below to guide yourself through the integration process:
Example 2.27. Sine Substitution.
Evaluate β«β1βx2dx.
Since \(\sqrt{1-x^2}\) appears in the integrand we try the trigonometric substitution \(x=\sin\theta\text{.}\) (Here we are using the restricted sine function with \(\theta\in[-\pi/2,~\pi/2]\) but typically omit this detail when writing out the solution.) Then \(dx=\cos \theta\,d\theta\text{.}\)
where, in the last line, we had \(|\cos \theta| = \cos \theta| \) since for \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\) we have \(\cos\theta\geq0\text{.}\)
Often we omit the step containing the absolute value by our discussion above. Now, to integrate a power of cosine we use the guideline for products of sine and cosine and make use of the identity
Our integral then becomes
To write the answer back in terms of \(x\) we use a right triangle. Since \(\sin\theta=x/1\) we have the triangle:
The triangle gives \(\sin\theta\text{,}\) \(\cos\theta\text{,}\) \(\tan\theta\text{,}\) but have a \(\sin(2\theta)\text{.}\) Thus, we use an identity to writeFor \(\theta\) by itself we use \(\theta=\sin^{-1}x\text{.}\) Thus, the integral is
Example 2.28. Secant Substitution.
Evaluate β«β25x2β4xdx.
We do not have \(\sqrt{x^2-a^2}\) because of the \(25\text{,}\) but if we factor \(25\) out we get:
Now, \(a=2/5\text{,}\) so let \(x=\frac{2}{5}\sec\theta\text{.}\) Alternatively, we can think of the integral as being:
Then we could let \(u=5x\) followed by \(u=2\sec\theta\text{,}\) etc. Or equivalently, we can avoid a \(u\)-substitution by letting \(5x=2\sec\theta\text{.}\) In either case we are using the trigonometric substitution \(x=\frac{2}{5}\sec\theta\text{,}\) but do use the method that makes the most sense to you! As \(x=\frac{2}{5}\sec\theta\) we have \(dx=\frac{2}{5}\sec\theta\tan\theta\,d\theta\text{.}\)
For \(\tan\theta\text{,}\) we use a right triangle.
Using SOH CAH TOA, the triangle is then
For \(\theta\) by itself, we use \(\theta=\sec^{-1}(5x/2)\text{.}\) Thus,Example 2.29. Tangent Substitution.
Evaluate β«1β25+x2dx.
Let \(x=5\tan\theta\) so that \(dx=5\sec^2\theta\,d\theta\text{.}\)
Since \(\tan\theta=x/5\text{,}\) we draw a triangle:
ThenTherefore, the integral is
Example 2.30. Completing the Square.
Evaluate β«xβ3β2xβx2dx.
First, complete the square to write
Now, we may let \(u=x+1\) so that \(du=dx\) (note that \(x=u-1\)) to get:
Let \(u=2\sin\theta\) giving \(du=2\cos\theta\,d\theta\text{:}\)
Integrating and using a triangle we get:
Note that in this problem we could have skipped the \(u\)-substitution if instead we let \(x+1=2\sin\theta\text{.}\) (For the triangle we would then use \(\ds\sin\theta=\frac{x+1}{2}\text{.}\))
Exercises for Section 2.3.
Exercise 2.3.1.
Evaluate the following indefinite integrals.
-
\(\ds\int\sqrt{x^2-1}\,dx\)
AnswerSolution\(\ds x\sqrt{x^2-1}/2-\ln|x+\sqrt{x^2-1}|/2+C\)We use the substitution:
\begin{equation*} x = \sec\theta,\ dx = \sec\theta\tan\theta\,d\theta\text{.} \end{equation*}Then:
\begin{equation*} \begin{split} \int \sqrt{x^2-1} \,dx \amp = \int \sqrt{\sec^2\theta-1}\,\sec\theta\tan\theta\,d\theta\\ \amp = \int \tan^2\theta \sec\theta\,d\theta\\ \amp = \int \sec^3\theta - \sec\theta \,d\theta\\ \amp = \frac{1}{2} \sec\theta\tan\theta + \frac{1}{2}\ln|\tan\theta+\sec\theta|-\ln|\sec\theta + \tan\theta+C\\ \amp = \frac{1}{2} \sec\theta\tan\theta -\frac{1}{2}\ln|\tan\theta+\sec\theta|+C. \end{split} \end{equation*}We now need to write the solution in terms of \(x\text{.}\) Since \(\sec\theta = x\text{,}\) this means that \(\tan\theta = \sqrt{x^2-1}\text{.}\) Thus:
\begin{equation*} \int \sqrt{x^2-1} = \frac{1}{2}x\sqrt{x^2-1} -\frac{1}{2}\ln|\sqrt{x^2-1}+x|+C\text{.} \end{equation*} -
\(\ds\int\sqrt{9+4x^2}\,dx\)
AnswerSolution\(\ds x\sqrt{9+4x^2}/2+\hbox{\)\ds(9/4)\ln|2x+\sqrt{9+4x^2}|+C\(}\)We first rewrite the integrand as
\begin{equation*} \sqrt{9+4x^2} = 2\sqrt{\frac{9}{4}+x^2}\text{.} \end{equation*}Now let
\begin{equation*} x=\frac{3}{2}\tan\theta,\ dx = \frac{3}{2}\sec^2\theta\,d\theta\text{.} \end{equation*}Then the integral becomes
\begin{equation*} \int 2\sqrt{\frac{9}{4}+x^2} \,dx = 2\int \frac{9}{4}\sec^3\theta\,d\theta\text{.} \end{equation*}Therefore, we find that
\begin{equation*} \int \sqrt{9+4x^2}\,dx = \frac{9}{4} \sec\theta\tan\theta + \frac{9}{4}\ln|\tan\theta+\sec\theta|+C\text{.} \end{equation*}If \(x= \frac{3}{2}\tan\theta\text{,}\) then we must have \(\sec\theta = \sqrt{1+4x^2/9}\text{.}\) Therefore, we find that
\begin{equation*} \int \sqrt{9+4x^2}\,dx = \frac{2x}{3}\sqrt{4x^2/9+1} + \frac{9}{4}\ln\left(\sqrt{\frac{4x^2}{9}+1}+\frac{2x}{3}\right)+C \end{equation*} -
\(\ds\int x\sqrt{1-x^2}\,dx\)
AnswerSolution\(\ds -(1-x^2)^{3/2}/3+C\)Let \(u=1-x^2\) with \(du = -2x\,dx\text{.}\) Then:
\begin{equation*} \begin{split} \int x\sqrt{1-x^2}\,dx \amp = \frac{-1}{2}\int \sqrt{u}\,du\\ \amp = \frac{-1}{3} u^{3/2} + C\\ \amp = \frac{-1}{3} \left(1-x^2\right)^{3/2}. \end{split} \end{equation*} -
\(\ds\int x^2\sqrt{1-x^2}\,dx\)
AnswerSolution\(\arcsin(x)/8-\sin(4\arcsin x)/32+C\)We see that \(\sqrt{1-x^2}\) is in the integrand, and so try the substitution:
\begin{equation*} x = \sin\theta, \ dx = -\sin\theta\, d\theta \text{ for } \theta \in [-\pi/2,\pi/2]\text{.} \end{equation*}This gives
\begin{equation*} \begin{split} \int x^2\sqrt{1-x^2} \amp = \int \sin^2\theta \sqrt{\cos^2\theta} \cos\theta\, d\theta = \int \sin^2\theta |\cos\theta| \cos\theta\, d\theta \\ \amp = \int \sin^2\theta \cos^2\theta \, d\theta \end{split} \end{equation*}Now use the identities \(\cos^2\theta = \frac{1}{2} (1+\cos(2\theta))\) and \(\sin^2\theta = \frac{1}{2}(1-\cos(2\theta))\text{:}\)
\begin{equation*} \begin{split} \int \sin^2\theta \cos^2\theta \, d\theta \amp = \int \frac{1}{4}\left[(1-\cos (2\theta))(1+\cos(2\theta))\right] \, d\theta \\ \amp = \int \frac{1}{4} \left[ 1- \cos(2\theta) + \cos(2\theta) - \cos^2 (2\theta)\right]\, d\theta \\ \amp = \int \frac{1}{4} \left[ 1 - \frac{1}{2}\left(1+\cos(4\theta)\right)\right]\, d\theta \\ \amp = \frac{1}{8} \theta - \frac{1}{32} \sin(4\theta) + C \end{split} \end{equation*}Rewriting in terms of \(x\text{,}\) we find
\begin{equation*} \int x^2\sqrt{1-x^2}\, dx = \frac{1}{8} \arcsin x - \frac{1}{32} \sin (4\arcsin x) + \hat{C}\text{.} \end{equation*} -
\(\ds\int{1\over\sqrt{1+x^2}}\,dx\)
AnswerSolution\(\ds \ln|x+\sqrt{1+x^2}|+C\)We take \(x=\tan\theta\text{,}\) with \(dx = \sec^2\theta d\theta\text{.}\) Therefore,
\begin{equation*} \begin{split} \int \frac{1}{\sqrt{1+x^2}}\,dx \amp = \int \frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}} \, d\theta \\ \amp = \int \frac{\sec^2\theta}{\sqrt{\sec^2\theta}}d\theta \\ \amp = \int \sec\theta d\theta = \ln \lvert \sec\theta + \tan\theta \rvert + C. \end{split} \end{equation*}We know that \(\tan \theta = x\text{,}\) and so \(\sec \theta= \sqrt{x^2+1}\text{.}\) Therefore,
\begin{equation*} \int \frac{1}{\sqrt{1+x^2}}\,dx = \ln \lvert \sqrt{x^2+1} + x\rvert + \hat{C}\text{.} \end{equation*} -
\(\ds\int\sqrt{x^2+2x}\,dx\)
AnswerSolution\(\ds (x+1)\sqrt{x^2+2x}/2-\hbox{\)\ds\ln|x+1+\sqrt{x^2+2x}|/2+C\(}\)First notice that \(x^2+2x = (x+1)^2-1\text{.}\) So we first make the substitution,
\begin{equation*} u = x+1, \ du = dx \end{equation*}giving
\begin{equation*} \int \sqrt{x^2+2x} \, dx = \int \sqrt{(x+1)^2 - 1} \, dx = \int \sqrt{u^2-1} \, du\text{.} \end{equation*}We now make a secant substitution. Let \(u = \sec \theta\) with \(du = \sec\theta \tan\theta \,d\theta\text{:}\)
\begin{equation*} \begin{split} \int \sqrt{u^2-1} \, du \amp = \int \sqrt{\sec^2\theta - 1} \sec\theta \tan\theta \, d\theta \\ \amp = \int \tan^2\theta \sec\theta \, d\theta \\ \amp = \int\sec\theta (\sec^2\theta - 1) \, d\theta \\ \amp = \int (\sec^3\theta - \sec\theta) d\theta \\ \amp = \frac{1}{2} \sec\theta \tan\theta - \frac{1}{2} \ln |\sec\theta + \tan\theta| + C. \end{split} \end{equation*}Since \(\sec \theta=u\text{,}\) we must have that \(\tan\theta = \sqrt{u^2-1}\text{.}\) Therefore,
\begin{equation*} \begin{split} \int \sqrt{x^2+2x} \, dx \amp = \frac{1}{2} u \sqrt{u^2-1} - \frac{1}{2} \ln |u + \sqrt{u^2-1}| + \hat{C}\\ \amp = \frac{1}{2} (x-1) \sqrt{(x-1)^2-1} - \frac{1}{2} \ln |(x-1) + \sqrt{(x-1)^2-1}| + \overline{C}, \end{split} \end{equation*}where we used \(x=u-1\text{.}\)
-
\(\ds\int{1\over x^2(1+x^2)}\,dx\)
AnswerSolution\(-\arctan x - 1/x+C\)We use the following substitution:
\begin{equation*} x = \tan\theta \text{ with } dx = \sec^2\theta\,d\theta \text{.} \end{equation*}Therefore, we have
\begin{equation*} \begin{split} \int \frac{1}{x^2(1+x^2)} \, dx \amp = \int \frac{1}{\tan^2\theta (1+\tan^2\theta) }\sec^2\theta\,d\theta \\ \amp = \int \frac{1}{\tan^2\theta \sec^2\theta }\sec^2\theta\,d\theta\\ \amp = \int \frac{1}{\tan^2\theta}\,d\theta\\ \amp = \int \cot^2 \theta \,d\theta \end{split} \end{equation*}We now use the identity \(\cot^2\theta = \csc^2\theta-1\text{:}\)
\begin{equation*} \begin{split} \int \cot^2 \theta \,d\theta \amp = \int \csc^2\theta \,d\theta - \int \,d\theta\\ \amp = -\cot\theta - \theta + C. \end{split} \end{equation*}Since \(x=\tan\theta\text{,}\) we have
\begin{equation*} \theta = \arctan x\text{,} \end{equation*}and
\begin{equation*} \cot\theta = \frac{1}{x}\text{.} \end{equation*}We find that
\begin{equation*} \int \frac{1}{x^2(1+x^2)} \, dx = -\frac{1}{x} - \arctan x + C\text{.} \end{equation*} -
\(\ds\int{x^2\over\sqrt{4-x^2}}\,dx\)
AnswerSolution\(\ds 2\arcsin(x/2)-x\sqrt{4-x^2}/2+C\)Let
\begin{equation*} x = 2\sin\theta, \text{ with } dx = 2\cos\theta\,d\theta\text{.} \end{equation*}Then the integral becomes
\begin{equation*} \begin{split} \int \frac{x^2}{\sqrt{4-x^2}} \, dx \amp = \int \frac{4 \sin^2\theta}{\sqrt{4-4\sin^2\theta}} \left(2 \cos\theta\right)\,d\theta\\ \amp = \int \frac{4 \sin^2\theta}{2\cos\theta} \left(2 \cos\theta\right)\,d\theta\\ \amp =4\int \sin^2\theta\,d\theta \end{split} \end{equation*}Now to evaluate this integral, we let
\begin{equation*} \sin^2 \theta = \frac{1}{2}(1-\cos(2\theta))\text{.} \end{equation*}Therefore,
\begin{equation*} 4\int \sin^2 \theta \,d\theta = 2 \int 1-\cos (2\theta) \,d\theta\text{.} \end{equation*}Now let \(u=2\theta\) with \(du = 2\,d\theta\text{:}\)
\begin{equation*} \begin{split} 2 \int 1-\cos (2\theta) \,d\theta \amp = \int (1-\cos(u))\,du\\ \amp = u - \sin u + C \end{split} \end{equation*}Hence,
\begin{equation*} 4 \int \sin^2 \theta \,d\theta = 2\theta - \sin (2\theta) + C\text{.} \end{equation*}Now since \(x = 2\sin\theta\text{,}\) we have
\begin{equation*} \theta = \arcsin\left(\frac{\theta}{2}\right)\text{,} \end{equation*}and
\begin{equation*} \sin (2\theta) = 2\sin\theta\cos\theta = x \frac{\sqrt{1-x^2}}{2}\text{.} \end{equation*}We have
\begin{equation*} \int \frac{x^2}{\sqrt{4-x^2}} \, dx= 2\arcsin\left(\frac{\theta}{2}\right) - \frac{x\sqrt{1-x^2}}{2} + C\text{.} \end{equation*} -
\(\ds\int{\sqrt{x}\over\sqrt{1-x}}\,dx\)
AnswerSolution\(\ds \arcsin(\sqrt{x})-\sqrt{x}\sqrt{1-x}+C\)We try the following substitution:
\begin{equation*} \sqrt{x} = \sin \theta \implies x = \sin^2\theta\text{,} \end{equation*}with
\begin{equation*} dx = 2\sin\theta\cos\theta\,d\theta\text{.} \end{equation*}Then the integral becomes
\begin{equation*} \begin{split} \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx \amp = \int \frac{\sin\theta}{\sqrt{1-\sin^2\theta}}2\sin\theta\cos\theta \,d\theta\\ \amp = \int \frac{\sin\theta}{\cos \theta} 2 \sin\theta\cos\theta\,d\theta\\ \amp = 2 \int \sin^2\theta\,d\theta. \end{split} \end{equation*}Now let
\begin{equation*} 2 \sin^2 \theta = \frac{1}{2}(1-\cos(2\theta))\text{.} \end{equation*}Therefore,
\begin{equation*} 2\int \sin^2 \theta \,d\theta = \int 1-\cos (2\theta) \,d\theta\text{.} \end{equation*}Now let \(u=2\theta\) with \(du = 2\,d\theta\text{:}\)
\begin{equation*} \begin{split} \int 1-\cos (2\theta) \,d\theta \amp = \frac{1}{4} \int (1-\cos(u))\,du\\ \amp =\frac{1}{2}\left( u - \sin u\right) + C \end{split} \end{equation*}Hence,
\begin{equation*} 2\int \sin^2 \theta \,d\theta = \theta - \frac{1}{2} \sin (2\theta) + C\text{.} \end{equation*}Recall that \(\sin(2\theta) = 2\sin\theta\cos\theta\text{.}\) Therefore:
\begin{equation*} 2\int \sin^2 \theta \,d\theta = \theta -\sin\theta\cos\theta + C\text{.} \end{equation*}Now we use the fact that
\begin{equation*} \theta = \arcsin \sqrt{x}\text{,} \end{equation*}and
\begin{equation*} \sin\theta\cos\theta = \sqrt{x} \sqrt{1-x} \end{equation*}to write
\begin{equation*} \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx = \arcsin\sqrt{x}- \sqrt{x}\sqrt{1-x} + C\text{.} \end{equation*} -
\(\ds\int{x^3\over\sqrt{4x^2-1}}\,dx\)
AnswerSolution\(\ds (2x^2+1)\sqrt{4x^2-1}/24+C\)We first rewrite the integrand:
\begin{equation*} \frac{x^3}{\sqrt{4x^2-1}} = \frac{x^3}{2\sqrt{x^2-(1/2)^2}}\text{.} \end{equation*}Therefore, we use the substitution:
\begin{equation*} x = \frac{\sec\theta}{2}, \text{ with } dx = \frac{\sec\theta\tan\theta}{2}\,d\theta\text{.} \end{equation*}Therefore the integral becomes:
\begin{equation*} \begin{split} \int \frac{x^3}{2\sqrt{x^2-1/4}}\,dx \amp = \int \frac{\sec^3\theta/8}{2\sqrt{(\sec^2\theta-1)/4}} \frac{\sec\theta\tan\theta}{2}\,d\theta\\ \amp = \int \frac{\sec^3\theta/8}{\tan\theta} \frac{\sec\theta\tan\theta}{2}\,d\theta \\ \amp = \frac{1}{16} \int \sec^4\theta \,d\theta \\ \amp = \frac{1}{16(3)} \bigl(\cos(2\theta) + 2\bigr) \tan\theta \sec^2 \theta + C \\ \end{split} \end{equation*}We now write the integral back in terms of \(x\text{:}\) since
we have\begin{equation*} \begin{split} \sec^2\theta \amp = 4x^2\\ \tan\theta \amp= \sqrt{4x^2-1}\\ \cos(2\theta) \amp = \frac{1-\sqrt{4x^2-1}}{4x^2} \end{split} \end{equation*}All together, we find that
\begin{equation*} \ds\int{x^3\over\sqrt{4x^2-1}}\,dx = \frac{1}{24} (2x^2+1)\sqrt{4x^2-1}+C \end{equation*}