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.

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.

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!

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 $\sqrt{x^{2}-a^{2}}$.

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.

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.

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.