OBJECTIVES

When you finish this section, you should be able to:

1. Find integrals containing \(\sqrt{ a^{2}- x^{ 2}}\) (p. 488)
2. Find integrals containing \(\sqrt{ x^{ 2} + a^{ 2}}\) (p. 489)
3. Find integrals containing \(\sqrt{ x^{ 2}- a^{ 2}}\) (p. 491)
4. Use trigonometric substitution to find definite integrals (p. 492)

CAUTION

Be careful to use the restrictions on each substitution. They guarantee the substitution is a one-to-one function, which is a requirement for using substitution.

NEED TO REVIEW?

Right triangle trigonometry is discussed in Appendix A.4, pp. A-27 to A-31.

Finding an Integral Containing \(\sqrt{4-x^{2}}\)

Find \(\int \dfrac{dx}{x^{2}\sqrt{4-x^{2}}}\).

Solution The integrand contains the square root \(\sqrt{4-x^{2}}\) that is of the form \(\sqrt{a^{2}-x^{2}}\), where \(a=2\). We use the substitution \(x=2\sin \theta\), \(-\dfrac{\pi }{2} \lt\theta \lt\dfrac{\pi }{2}\). Then \(dx=2\cos \theta \,d\theta\). Since \[ \begin{eqnarray*} {\sqrt{4-x^{2}}} \underset{\underset{\color{#0066A7}{\hbox{\(x=2\sin \theta\)}}}{\color{#0066A7}{\uparrow }}} {=} {\sqrt{4-4\sin ^{2}\theta }}=2{\sqrt{1-\sin ^{2}\theta }}=2{\sqrt{\cos ^{2}\theta }} \underset{\underset{\color{#0066A7}{\hbox{\(\text {cos}~\theta \gt 0~\text {since} -\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2}\)}}}{\color{#0066A7}{\uparrow}}}{=}2\cos \theta \end{eqnarray*} \]

we have \[ \int\! \frac{dx}{x^{2}\sqrt{4-x^{2}}}=\int\! \frac{2\cos \theta \,d\theta }{ (4\sin ^{2}\theta )(2\cos \theta )}=\int\! \dfrac{d\theta }{4\sin ^{2}\theta }= \frac{1}{4}\int \csc ^{2}\theta \,d\theta =-\frac{1}{4}\cot \theta +C \]

The original integral is a function of \(x\), but the solution above is a function of \(\theta\). To express \(\cot \theta\) in terms of \(x\), refer to the right triangles drawn in Figure 4.

Figure 4 \(\sin \theta =\dfrac{x}{2}\), \(-\dfrac {\pi}{2} \lt \theta \lt \dfrac {\pi}{2}\)

Using the Pythagorean Theorem, the third side of each triangle is \(\sqrt{ 2^{2}-x^{2}}=\sqrt{4-x^{2}}\). So, \[ \cot \theta =\dfrac{\sqrt{4-x^{2}}}{x} \qquad -\dfrac{\pi }{2} \lt\theta \lt \dfrac{\pi }{2} \]

Then \[ \int \dfrac{dx}{x^{2}\sqrt{4-x^{2}}}=-\frac{1}{4}\cot \theta +C=-\frac{1}{4}\frac{\sqrt{4-x^{2}}}{x}+C=-\frac{\sqrt{4-x^{2}}}{4x}+C \]

NOW WORK

Problem 7.

Finding an Integral Containing \(\sqrt{x^{2}+9}\)

Find \(\int \frac{dx}{(x^{2}+9)^{3/2}}\).

Solution The integral contains a square root \((x^{2}+9)^{3/2}= \big( \sqrt{x^{2}+9}\big) ^{3}\) that is of the form \(\sqrt{x^{2}+a^{2}},\) where \(a=3\). We use the substitution \(x=3\tan \theta\), \(-\dfrac{\pi }{2} \lt \theta \lt \dfrac{\pi }{2}\). Then \(dx=3\sec ^{2}\theta d\theta\). Since \[ \begin{eqnarray*} (x^{2}+9)^{3/2}\underset{\underset{\color{#0066A7}{\hbox{\(x=3\tan \theta\)}}}{\color{#0066A7}{\uparrow}}}{=}(9\tan^{2}\theta +9)^{3/2}=9^{3/2}(\tan^{2}\theta +1)^{3/2}\underset{\underset{\color{#0066A7}{\hbox{\(\tan^{2}\theta +1=\sec ^{2}\theta\)}}}{\color{#0066A7}{\uparrow}}}{=}27(\sec^{2}\theta)^{3/2}\underset{\underset{\color{#0066A7}{\hbox{\(\sec \theta >0\)}}}{\color{#0066A7}{\uparrow}}}{=}27\sec ^{3}\theta \end{eqnarray*} \]

Figure 5 \(\tan \theta =\dfrac{x}{3},\) \(-\dfrac{\pi }{2} \lt \theta \lt \dfrac{\pi }{2}\)

we have \[ \begin{equation*} \int \frac{dx}{(x^{2}+9)^{3/2}}=\int \frac{3\sec ^{2}\theta d\theta }{27\sec ^{3}\theta }=\frac{1}{9}\int \frac{d\theta }{ \sec \theta }=\frac{1}{9}\int \cos \theta d\theta =\frac{1}{9}\sin \theta +C \end{equation*} \]

To express the solution in terms of \(x,\) use either the right triangles in Figure 5 or identities.

From the right triangles, the hypotenuse is \(\sqrt{x^{2}+3^{2}}=\sqrt{x^{2}+9 }\). So, \(\sin \theta =\dfrac{x}{\sqrt{x^{2}+9}}.\) Then \[ \int \frac{dx}{(x^{2}+9)^{3/2}}=\dfrac{1}{9}\sin \theta +C=\dfrac{x}{9 \sqrt{x^{2}+9}}+C \]

NOW WORK

Problem 15.

NOTE

An integral containing \(\sqrt{x^{2}+a^{2}}\), \(a \gt 0\), can also be found using the substitution \(x=a \sinh \theta\), since \[ \begin{eqnarray*} \sqrt{x^{2}+a^2} &=& \sqrt{a^2\sinh ^2 \theta +a^2} \\[2pt] &=& a \sqrt{\sinh ^{2} \theta +1} \\[2pt] &=&a \sqrt{\cosh ^{2} \theta} = a \cosh \theta \end{eqnarray*} \]

Try it!

Finding the Integral \(\int (4x^{2}+9) ^{1/2}dx\)

Find \(\int ( 4x^{2}+9) ^{1/2}dx\).

Solution \(\int (4x^{2}+9)^{1/2}dx =\int \sqrt{(2x)^{2}+3^{2}} dx\)

We use the substitution \(2x=3\tan \theta\), \(-\dfrac{\pi }{2} \lt \theta \lt \dfrac{\pi}{2}.\) Then \(dx=\dfrac{3}{2}\sec ^{2}\theta d\theta\) and \[ \begin{eqnarray*} \int ( 4x^{2}+9) ^{1/2}dx &=&\dfrac{3}{2}\int \sqrt{9\tan ^{2}\theta +9}\sec ^{2}\theta\, d\theta =\dfrac{9}{2}\int \sqrt{\tan ^{2}\theta +1}\sec ^{2}\theta\, d\theta \\[5pt] &=&\dfrac{9}{2}\int \sec ^{3}\theta \,d\theta \\[5pt] &=&\dfrac{9}{2}\left[ \dfrac{1}{2}\sec \theta \tan \theta +\dfrac{1}{2}\ln \left\vert \sec \theta +\tan \theta \right\vert \right] +C \end{eqnarray*} \]

Figure 6 \(\tan \theta =\dfrac{2x}{3}\), \(-\dfrac{\pi }{2} \lt \theta \lt \dfrac{\pi }{2}\)

To express the solution in terms of \(x\), refer to the right triangles drawn in Figure 6.

Using the Pythagorean Theorem, the hypotenuse of each triangle is \(\sqrt{ (2x) ^{2}+9}=\sqrt{4x^{2}+9}.\) So, \[ \sec \theta =\frac{\sqrt{4x^{2}+9}}{3}\quad \hbox{and}\quad \tan \theta =\frac{2x}{3} \quad -\frac{\pi }{2} \lt \theta \lt \frac{\pi }{2} \]

Then \[ \begin{eqnarray*} \int ( 4x^{2}+9) ^{1/2}dx &=&\frac{9}{4}\left[ \sec \theta \tan \theta +\ln \left\vert \sec \theta +\tan \theta \right\vert \right] +C \nonumber \\[2pt] &=&\frac{9}{4}\left[ \frac{\sqrt{4x^{2}+9}}{3}\cdot \frac{2x}{3}+\ln \left\vert \frac{\sqrt{4x^{2}+9}}{3}+\frac{2x}{3}\right\vert \right] +C \nonumber \\[3pt] &=&\frac{9}{4}\left[ \frac{2x\sqrt{4x^{2}+9}}{9}+\ln \frac{2x+\sqrt{4x^{2}+9}}{3}\right] +C \end{eqnarray*} \]

NOW WORK

Problem 45.

NOTE

The substitution \(x=a\cosh \theta \) can also be used for integrands containing \(\sqrt{x^{2}-a^{2}}\).

Finding an Integral Containing \(\sqrt{x^{2}-4}\)

Find \(\int \frac{\sqrt{x^{2}-4}}{x} dx\).

Solution The integrand contains the square root \(\sqrt{x^{2}-4}\) that is of the form \(\sqrt{x^{2}-a^{2}},\) where \(a=2\). We use the substitution \(x=2\sec \theta\), \(0 \le \theta \lt \dfrac{\pi }{2},\) \(\pi \le \theta \lt\dfrac{3\pi }{2}\). Then \(dx=2\sec \theta \tan \theta~d\theta.\) Since \[ \begin{eqnarray*} \sqrt{x^{2}-4} \underset{\underset{\color{#0066A7}{\hbox{\(x=2 \sec \theta\)}}}{\color{#0066A7}{\uparrow}}}{=}\sqrt{4\sec ^{2}\theta -4}=2\sqrt{\sec ^{2}\theta -1}=2\sqrt{\tan^{2}\theta } \underset{\underset{\underset{\color{#0066A7}{\hbox{since \(0 \le \theta \lt \dfrac{\pi }{2}, \pi \le \theta \lt \dfrac{3\pi}{2}\)}}}{\color{#0066A7}{\hbox{\(\tan \theta \ge 0\)}}}}{\color{#0066A7}{\uparrow}}}{=} 2\tan \theta \end{eqnarray*} \]

we have \[ \begin{eqnarray*} \int \frac{\sqrt{x^{2}-4}}{x}dx&=&\int \frac{(2\tan \theta )(2\sec \theta \tan \theta \,d\theta )}{2\sec \theta }=2\int \tan ^{2}\theta d\theta \underset{\underset{\color{#0066A7}{\hbox{\(\tan^{2}\theta = \sec^{2}\theta -1\)}}}{\color{#0066A7}{\uparrow}}}{=}2\int (\sec^{2}\theta -1) d\theta \\ &=&2\int \sec^{2}\theta d\theta -2\int d\theta =2\tan \theta -2\theta +C \end{eqnarray*} \]

To express the solution in terms of \(x\), we use either the right triangles in Figure 7 or trigonometric identities.

Using identities, we find, \[ \begin{eqnarray*} \tan \theta \underset{\underset{\underset{{\color{#0066A7}{\hbox{\(\tan \theta \ge 0\)}}}}{\color{#0066A7}{\hbox{\(\tan^{2}\theta =\sec^{2}\theta -1\)}}}}{\color{#0066A7}{\uparrow}}}{=}\sqrt{\sec^{2}\theta -1}= \sqrt{\frac{x^{2}}{4}-1}=\frac{1}{2}\sqrt{x^{2}-4} \end{eqnarray*} \]

Also since \(\sec \theta =\dfrac{x}{2},\) \(0 \le \theta \lt \dfrac{\pi }{2},\) \(\pi \le \theta \lt \dfrac{3\pi }{2},\) the inverse function \(\theta =\sec ^{-1}\dfrac{x }{2}\) is defined.

Then \[ \int\frac{\sqrt{x^{2}-4}}{x}dx=2\tan \theta -2\theta+C=\sqrt{x^{2}-4} -2\sec ^{-1} \frac{x}{2}+C \]

NOW WORK

Problem 29.

Finding the Area Enclosed by an Ellipse

Find the area \(A\) enclosed by the ellipse \(\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1.\)

Figure 8 \(\,\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1\)

Solution Figure 8 illustrates the ellipse. Since the ellipse is symmetric with respect to both the \(x\)-axis and the \(y\)-axis, the total area \(A\) of the ellipse is four times the shaded area in the first quadrant, where \(0 \le x \le 2\) and \(0 \le y \le 3.\)

We begin by expressing \(y\) as a function of \(x.\) \[ \begin{eqnarray*} \frac{x^{2}}{4}+\frac{y^{2}}{9} &=&1 \\[4.5pt] \frac{y^{2}}{9} &=&1-\frac{x^{2}}{4}=\frac{4-x^{2}}{4} \\[4.5pt] y^{2} &=&\dfrac{9}{4}( 4-x^{2}) \\[4.5pt] y &=&\dfrac{3}{2}\sqrt{4-x^{2}}\qquad {\color{#0066A7}{\hbox{\(y \ge 0\)}}} \end{eqnarray*} \]

So, the area \(A\) of the ellipse is four times the area under the graph of \(y=\dfrac{3}{2}\sqrt{4-x^{2}},\) \(0 \le x \le 2.\) That is, \[ A=4 \int_{0}^{2}\dfrac{3}{2}\sqrt{4-x^{2}}\,dx =6\int_{0}^{2}\sqrt{4-x^{2}}\,dx \]

Since the integrand contains a square root of the form \(\sqrt{a^{2}-x^{2}}\) with \(a=2,\) we use the substitution \(x=2\sin \theta\), \(-\frac{\pi}{2} \le \theta \le \frac{\pi }{2}\). Then \(dx=2\cos \theta \,d\theta\). The new limits of integration are:

• When \(x=0\), \(2\sin \theta =0\), so \(\theta =0\).
• When \(x=2\), \(2\sin \theta =2\), so \(\sin \theta =1\) and \(\theta =\dfrac{\pi }{2}\).

Then \[ \begin{eqnarray*} A&=&6\int_{0}^{2}\sqrt{4-x^{2}}\,dx=6\int_{0}^{\pi /2}\sqrt{4-4\sin^{2}\theta }\cdot 2\cos \theta \,d\theta\\ &=&6\int_{0}^{\pi /2}2\sqrt{1-\sin^{2}\theta }\cdot 2\cos \theta \,d\theta = 24\int_{0}^{\pi /2}\sqrt{\cos ^{2}\theta }\cdot \cos \theta \,d\theta\nonumber \\ &\underset{\underset{\color{#0066A7}{\hbox{\(\cos \theta \ge 0\)}}}{\color{#0066A7}{\uparrow }}}{=}&24\int_{0}^{\pi /2}\cos^{2}\theta \,d\theta \underset{\underset{\color{#0066A7}{\hbox{\(\cos^{2}\theta =\frac{1+\cos (2\theta )}{2}\)}}}{\color{#0066A7}{\uparrow }}}{=} \frac{24}{2}\int_{0}^{\pi /2}\left[1+\cos (2\theta ) \right] \,d\theta \nonumber \\ & =&12\left[ \theta +\frac{1}{2}\sin (2\theta ) \right]_{0}^{\pi /2}=12\left( \frac{\pi }{2}+0\right) =6\pi \end{eqnarray*} \]

The area of the ellipse is \(6\pi\) square units.

NOW WORK

Problem 63.

NEED TO REVIEW?

The two approaches to finding a definite integral using the method of substitution are discussed in Section 5.6, pp. 387-393.

Use Trigonometric Substitution to Find a Definite Integral

Find the area under the graph of \(y=\sqrt{x^{2}-1}\) (the upper half of the right branch of the hyperbola \(y^{2}=x^{2}-1\)) from 1 to 3. See Figure 9.

Figure 9 \(y=\sqrt{x^{2}-1}\), \(1 \le x \le 3\)

Solution The area \(A\) we seek is \(A= \int_{1}^{3}\sqrt{x^{2}-1 }\,dx\). The integral contains a square root of the form \(\sqrt{x^{2}-a^{2}}\), where \(a=1\), so we use the trigonometric substitution \(x=\sec \theta \), \( 0 \le \theta \lt\dfrac{\pi }{2},\) \(\pi \le \theta \lt\dfrac{3\pi }{2}.\) Then \(dx=\sec \theta \tan \theta \,d\theta \). Since the upper limit \(x=3\) does not result in a nice angle \((\theta =\sec ^{-1}\,3)\), we find the indefinite integral first and then use the Fundamental Theorem of Calculus.

With \(x=\sec \theta \hbox{ and } dx=\sec \theta \tan \theta d\theta ,\) we have \[ \begin{eqnarray*} A&=&\int \sqrt{x^{2}-1}\,dx=\int \sqrt{\sec ^{2}\theta -1}\sec \theta \tan \theta \,d\theta =\int \tan \theta \cdot \sec \theta \tan \theta \,d\theta\\[5pt] &=&\int \tan ^{2}\theta \sec \theta d\theta \end{eqnarray*} \]

Since \(\tan \theta \) is raised to an even power and \(\sec \theta \) to an odd power, we use the identity \(\tan ^{2}\theta =\sec ^{2}\theta -1.\)Then \[ \begin{eqnarray*} A &=&\int \sqrt{x^{2}-1}\,dx=\int \tan ^{2}\theta \sec \theta d\theta =\int \left( \sec ^{2}\theta -1\right) \sec \theta d\theta\\[4pt] &=&\int \sec ^{3}\theta d\theta -\int \sec \theta d\theta \\[4pt] &=&\dfrac{1}{2}\left[ \sec \theta \tan \theta +\ln \left\vert \sec \theta +\tan \theta \right\vert \right] -\ln \left\vert \sec \theta +\tan \theta \right\vert +C\\[4pt] &=&\dfrac{1}{2}\sec \theta \tan \theta -\dfrac{1}{2}\ln \left\vert \sec \theta +\tan \theta \right\vert +C \end{eqnarray*} \]

Now we express \(\tan \theta \) in terms of \(x=\sec \theta ,\) and apply the Second Fundamental Theorem of Calculus. \[ \begin{eqnarray*} A&=&\int_{1}^{3}\sqrt{x^{2}-1}\, dx\underset{\underset{\underset{\color{#0066A7}{\hbox{\(\tan \theta =\sqrt{x^{2} -1}\)}}}{\color{#0066A7}{\hbox{\(\sec \theta =x\)}}}}{\color{#0066A7}{\uparrow }}} {=}\left[ \frac{1}{2}x\sqrt{x^{2}-1}-\frac{1}{2}\ln \left\vert x+ \sqrt{x^{2}-1}\right\vert \right] _{1}^{3}\\ &=&\dfrac{3}{2}\sqrt{8}-\dfrac{1}{2}\ln ( 3+\sqrt{8}) =3\sqrt{2} -\dfrac{1}{2}\ln (3+2\sqrt{2}) \end{eqnarray*} \]

NOW WORK

Problem 53.