Trigonometric Substitution

In this section, we explore integrals containing expressions of the form

\sqrt{a^{2} - x^{2}},

are positive. We have already encountered and evaluated integrals containing some expressions of this type, but many still remain inaccessible. The technique of trigonometric substitution comes in very handy when evaluating these integrals. This technique uses substitution to rewrite these integrals as trigonometric integrals.

Integrals Involving

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

Before developing a general strategy for integrals containing

\sqrt{a^{2} - x^{2}},

This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution

x = 3 sin θ,

At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions. Before completing this example, let’s take a look at the general theory behind this idea.

To see that this actually makes sense, consider the following argument: The domain of

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

We can see, from this discussion, that by making the substitution

x = a sin θ,

we are able to convert an integral involving a radical into an integral involving trigonometric functions. After we evaluate the integral, we can convert the solution back to an expression involving

x .

we can draw the reference triangle in [link] to assist in expressing the values of

cos θ,

It can be shown that this triangle actually produces the correct values of the trigonometric functions evaluated at

θ

The essential part of this discussion is summarized in the following problem-solving strategy.

The following example demonstrates the application of this problem-solving strategy.

Integrating an Expression Involving

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

Evaluate $\int^{} \sqrt{9 - x^{2}} d x .$

Begin by making the substitutions $x = 3 sin θ$

and $d x = 3 cos θ d θ .$

Since $sin θ = \frac{x}{3},$

we can construct the reference triangle shown in the following figure.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled 3, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (9 – x^2). To the left of the triangle is the equation sin(theta) = x/3.

Thus,

\begin{matrix} \int^{​} \sqrt{9 - x^{2}} d x & = \int^{​} \sqrt{9 - {(3 sin θ)}^{2}} 3 cos θ d θ & Substitute x = 3 sin θ and d x = 3 cos θ d θ . \\ = \int^{​} \sqrt{9 (1 - {sin}^{2} θ)} 3 cos θ d θ & Simplify. \\ = \int^{​} \sqrt{9 {cos}^{2} θ} 3 cos θ d θ & Substitute {cos}^{2} θ = 1 - {sin}^{2} θ . \\ = \int^{​} 3 \| cos θ \| 3 cos θ d θ & Take the square root. \\ = \int^{​} 9 {cos}^{2} θ d θ & \begin{matrix} Simplify. Since - \frac{π}{2} \leq θ \leq \frac{π}{2}, cos θ \geq 0 and \\ \| cos θ \| = cos θ . \end{matrix} \\ = \int^{​} 9 (\frac{1}{2} + \frac{1}{2} cos (2 θ)) d θ & \begin{matrix} Use the strategy for integrating an even power \\ of cos θ . \end{matrix} \\ = \frac{9}{2} θ + \frac{9}{4} sin (2 θ) + C & Evaluate the integral. \\ = \frac{9}{2} θ + \frac{9}{4} (2 sin θ cos θ) + C & Substitute sin (2 θ) = 2 sin θ cos θ . \\ = \frac{9}{2} {sin}^{−1} (\frac{x}{3}) + \frac{9}{2} \cdot \frac{x}{3} \cdot \frac{\sqrt{9 - x^{2}}}{3} + C & \begin{matrix} Substitute {sin}^{−1} (\frac{x}{3}) = θ and sin θ = \frac{x}{3} . Use \\ the reference triangle to see that \\ cos θ = \frac{\sqrt{9 - x^{2}}}{3} and make this substitution. \end{matrix} \\ = \frac{9}{2} {sin}^{−1} (\frac{x}{3}) + \frac{x \sqrt{9 - x^{2}}}{2} + C . & Simplify. \end{matrix}

Integrating an Expression Involving

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

Two Ways

Evaluate $\int^{} x^{3} \sqrt{1 - x^{2}} d x$

two ways: first by using the substitution $u = 1 - x^{2}$

and then by using a trigonometric substitution.

Method 1

Let $u = 1 - x^{2}$

and hence $x^{2} = 1 - u .$

Thus, $d u = −2 x d x .$

In this case, the integral becomes

\begin{matrix} \int^{​} x^{3} \sqrt{1 - x^{2}} d x & = - \frac{1}{2} \int^{​} x^{2} \sqrt{1 - x^{2}} (−2 x d x) & Make the substitution. \\ = - \frac{1}{2} \int^{​} (1 - u) \sqrt{u} d u & Expand the expression. \\ = - \frac{1}{2} \int (u^{1 / 2} - u^{3 / 2}) d u & Evaluate the integral. \\ = - \frac{1}{2} (\frac{2}{3} u^{3 / 2} - \frac{2}{5} u^{5 / 2}) + C & Rewrite in terms of x . \\ = - \frac{1}{3} {(1 - x^{2})}^{3 / 2} + \frac{1}{5} {(1 - x^{2})}^{5 / 2} + C . \end{matrix}

Method 2

Let $x = sin θ .$

In this case, $d x = cos θ d θ .$

Using this substitution, we have

\begin{matrix} \int^{​} x^{3} \sqrt{1 - x^{2}} d x & = \int^{​} {sin}^{3} θ {cos}^{2} θ d θ \\ = \int^{​} (1 - {cos}^{2} θ) {cos}^{2} θ sin θ d θ & Let u = cos θ . Thus, d u = − sin θ d θ . \\ = \int^{​} (u^{4} - u^{2}) d u \\ = \frac{1}{5} u^{5} - \frac{1}{3} u^{3} + C & Substitute cos θ = u . \\ = \frac{1}{5} {cos}^{5} θ - \frac{1}{3} {cos}^{3} θ + C & \begin{matrix} Use a reference triangle to see that \\ cos θ = \sqrt{1 - x^{2}} . \end{matrix} \\ = \frac{1}{5} {(1 - x^{2})}^{5 / 2} - \frac{1}{3} {(1 - x^{2})}^{3 / 2} + C . \end{matrix}

Integrating Expressions Involving

\sqrt{a^{2} + x^{2}}

let’s first consider the domain of this expression. Since

\sqrt{a^{2} + x^{2}}

we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our choice is restricted to selecting either

x = a tan θ

Either of these substitutions would actually work, but the standard substitution is

x = a tan θ

With this substitution, we make the assumption that

− (π / 2) < θ < π / 2,

The procedure for using this substitution is outlined in the following problem-solving strategy.

Integrating an Expression Involving

\sqrt{a^{2} + x^{2}}

Evaluate $\int \frac{d x}{\sqrt{1 + x^{2}}}$

and check the solution by differentiating.

Begin with the substitution $x = tan θ$

and $d x = {sec}^{2} θ d θ .$

Since $tan θ = x,$

draw the reference triangle in the following figure.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (1+x^2), the vertical leg is labeled x, and the horizontal leg is labeled 1. To the left of the triangle is the equation tan(theta) = x/1.

Thus,

\begin{matrix} \int \frac{d x}{\sqrt{1 + x^{2}}} & = \int \frac{{sec}^{2} θ}{sec θ} d θ & \begin{matrix} Substitute x = tan θ and d x = {sec}^{2} θ d θ . This \\ substitution makes \sqrt{1 + x^{2}} = sec θ . Simplify. \end{matrix} \\ = \int^{​} sec θ d θ & Evaluate the integral. \\ = ln \| sec θ + tan θ \| + C & \begin{matrix} Use the reference triangle to express the result \\ in terms of x . \end{matrix} \\ = ln \| \sqrt{1 + x^{2}} + x \| + C . \end{matrix}

To check the solution, differentiate:

\begin{matrix} \frac{d}{d x} (ln \| \sqrt{1 + x^{2}} + x \|) & = \frac{1}{\sqrt{1 + x^{2}} + x} \cdot (\frac{x}{\sqrt{1 + x^{2}}} + 1) \\ = \frac{1}{\sqrt{1 + x^{2}} + x} \cdot \frac{x + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}} \\ = \frac{1}{\sqrt{1 + x^{2}}} . \end{matrix}

Since $\sqrt{1 + x^{2}} + x > 0$

for all values of $x,$

we could rewrite $ln \| \sqrt{1 + x^{2}} + x \| + C = ln (\sqrt{1 + x^{2}} + x) + C,$

if desired.

Evaluating

\int \frac{d x}{\sqrt{1 + x^{2}}}

Using a Different Substitution

Use the substitution $x = sinh θ$

to evaluate $\int \frac{d x}{\sqrt{1 + x^{2}}} .$

Because $sinh θ$

has a range of all real numbers, and $1 + {sinh}^{2} θ = {cosh}^{2} θ,$

we may also use the substitution $x = sinh θ$

to evaluate this integral. In this case, $d x = cosh θ d θ .$

Consequently,

\begin{matrix} \int \frac{d x}{\sqrt{1 + x^{2}}} & = \int \frac{cosh θ}{\sqrt{1 + {sinh}^{2} θ}} d θ & \begin{matrix} Substitute x = sinh θ and d x = cosh θ d θ . \\ Substitute 1 + {sinh}^{2} θ = {cosh}^{2} θ . \end{matrix} \\ = \int \frac{cosh θ}{\sqrt{{cosh}^{2} θ}} d θ & \sqrt{{cosh}^{2} θ} = \| cosh θ \| \\ = \int \frac{cosh θ}{\| cosh θ \|} d θ & \| cosh θ \| = cosh θ since cosh θ > 0 for all θ . \\ = \int \frac{cosh θ}{cosh θ} d θ & Simplify. \\ = \int^{​} 1 d θ & Evaluate the integral. \\ = θ + C & Since x = sinh θ, we know θ = {sinh}^{−1} x . \\ = {sinh}^{−1} x + C . \end{matrix}

Analysis

This answer looks quite different from the answer obtained using the substitution $x = tan θ .$

To see that the solutions are the same, set $y = {sinh}^{−1} x .$

Thus, $sinh y = x .$

From this equation we obtain:

\frac{e^{y} - e^{− y}}{2} = x .

After multiplying both sides by $2 e^{y}$

and rewriting, this equation becomes:

e^{2 y} - 2 x e^{y} - 1 = 0 .

Use the quadratic equation to solve for $e^{y} :$

e^{y} = \frac{2 x \pm \sqrt{4 x^{2} + 4}}{2} .

Simplifying, we have:

e^{y} = x \pm \sqrt{x^{2} + 1} .

Since $x - \sqrt{x^{2} + 1} < 0,$

it must be the case that $e^{y} = x + \sqrt{x^{2} + 1} .$

Thus,

y = ln (x + \sqrt{x^{2} + 1}) .

Last, we obtain

{sinh}^{−1} x = ln (x + \sqrt{x^{2} + 1}) .

After we make the final observation that, since $x + \sqrt{x^{2} + 1} > 0,$

ln (x + \sqrt{x^{2} + 1}) = ln \| \sqrt{1 + x^{2}} + x \|,

we see that the two different methods produced equivalent solutions.

Finding an Arc Length

Find the length of the curve $y = x^{2}$

over the interval $[0, \frac{1}{2}] .$

Because $\frac{d y}{d x} = 2 x,$

the arc length is given by

\int_{0}^{1 / 2} \sqrt{1 + {(2 x)}^{2}} d x = \int_{0}^{1 / 2} \sqrt{1 + 4 x^{2}} d x .

To evaluate this integral, use the substitution $x = \frac{1}{2} tan θ$

and $d x = \frac{1}{2} {sec}^{2} θ d θ .$

We also need to change the limits of integration. If $x = 0,$

then $θ = 0$

and if $x = \frac{1}{2},$

then $θ = \frac{π}{4} .$

Thus,

\begin{matrix} \int_{0}^{1 / 2} \sqrt{1 + 4 x^{2}} d x & = \int_{0}^{π / 4} \sqrt{1 + {tan}^{2} θ} \frac{1}{2} {sec}^{2} θ d θ & \begin{matrix} After substitution, \\ \sqrt{1 + 4 x^{2}} = tan θ . Substitute \\ 1 + {tan}^{2} θ = {sec}^{2} θ and simplify. \end{matrix} \\ = \frac{1}{2} \int_{0}^{π / 4} {sec}^{3} θ d θ & \begin{matrix} We derived this integral in the \\ previous section. \end{matrix} \\ = \frac{1}{2} (\frac{1}{2} sec θ tan θ + ln \| sec θ + tan θ \|) \|_{\begin{matrix} 0 \end{matrix}}^{\begin{matrix} π / 4 \end{matrix}} & Evaluate and simplify. \\ = \frac{1}{4} (\sqrt{2} + ln (\sqrt{2} + 1)) . \end{matrix}

Integrating Expressions Involving

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

The procedure for using this substitution is outlined in the following problem-solving strategy.

Finding the Area of a Region

Find the area of the region between the graph of $f (x) = \sqrt{x^{2} - 9}$

and the x-axis over the interval $[3, 5] .$

First, sketch a rough graph of the region described in the problem, as shown in the following figure.

This figure is the graph of the function f(x) = the square root of (x^2-9). It is an increasing curve that starts on the x-axis at 3 and is in the first quadrant. Under the curve above the x-axis is a shaded region bounded to the right at x = 5.

We can see that the area is $A = \int_{3}^{5} \sqrt{x^{2} - 9} d x .$

To evaluate this definite integral, substitute $x = 3 sec θ$

and $d x = 3 sec θ tan θ d θ .$

We must also change the limits of integration. If $x = 3,$

then $3 = 3 sec θ$

and hence $θ = 0 .$

If $x = 5,$

then $θ = {sec}^{−1} (\frac{5}{3}) .$

After making these substitutions and simplifying, we have

\begin{matrix} Area & = \int_{3}^{5} \sqrt{x^{2} - 9} d x \\ = \int_{0}^{{sec}^{−1} (5 / 3)} 9 {tan}^{2} θ sec θ d θ & Use {tan}^{2} θ = 1 - {sec}^{2} θ . \\ = \int_{0}^{{sec}^{−1} (5 / 3)} 9 ({sec}^{2} θ - 1) sec θ d θ & Expand. \\ = \int_{0}^{{sec}^{−1} (5 / 3)} 9 ({sec}^{3} θ - sec θ) d θ & Evaluate the integral. \\ = (\frac{9}{2} ln \| sec θ + tan θ \| + \frac{9}{2} sec θ tan θ) - 9 ln \| sec θ + tan θ \| \|_{\begin{matrix} 0 \end{matrix}}^{\begin{matrix} {sec}^{−1} (5 / 3) \end{matrix}} & Simplify. \\ = \frac{9}{2} sec θ tan θ - \frac{9}{2} ln \| sec θ + tan θ \| \|_{\begin{matrix} 0 \end{matrix}}^{\begin{matrix} {sec}^{−1} (5 / 3) \end{matrix}} & \begin{matrix} Evaluate. Use sec ({sec}^{−1} \frac{5}{3}) = \frac{5}{3} \\ and tan ({sec}^{−1} \frac{5}{3}) = \frac{4}{3} . \end{matrix} \\ = \frac{9}{2} \cdot \frac{5}{3} \cdot \frac{4}{3} - \frac{9}{2} ln \| \frac{5}{3} + \frac{4}{3} \| - (\frac{9}{2} \cdot 1 \cdot 0 - \frac{9}{2} ln \| 1 + 0 \|) \\ = 10 - \frac{9}{2} ln 3 . \end{matrix}

Key Concepts

Simplify the following expressions by writing each one using a single trigonometric function.

4 - 4 {sin}^{2} θ

9 {sec}^{2} θ - 9

9 {tan}^{2} θ

a^{2} + a^{2} {tan}^{2} θ

a^{2} + a^{2} {sinh}^{2} θ

a^{2} {cosh}^{2} θ

16 {cosh}^{2} θ - 16

Use the technique of completing the square to express each trinomial as the square of a binomial.

4 x^{2} - 4 x + 1

4 {(x - \frac{1}{2})}^{2}

2 x^{2} - 8 x + 3

− x^{2} - 2 x + 4

− {(x + 1)}^{2} + 5

Integrate using the method of trigonometric substitution. Express the final answer in terms of the variable.

\int \frac{d x}{\sqrt{4 - x^{2}}}

\int \frac{d x}{\sqrt{x^{2} - a^{2}}}

ln \| x + \sqrt{− a^{2} + x^{2}} \| + C

\int \sqrt{4 - x^{2}} d x

\int \frac{d x}{\sqrt{1 + 9 x^{2}}}

\frac{1}{3} ln \| \sqrt{9 x^{2} + 1} + 3 x \| + C

\int \frac{x^{2} d x}{\sqrt{1 - x^{2}}}

\int \frac{d x}{x^{2} \sqrt{1 - x^{2}}}

- \frac{\sqrt{1 - x^{2}}}{x} + C

\int \frac{d x}{{(1 + x^{2})}^{2}}

\int \sqrt{x^{2} + 9} d x

9 [\frac{x \sqrt{x^{2} + 9}}{18} + \frac{1}{2} l n \| \frac{\sqrt{x^{2} + 9}}{3} + \frac{x}{3} \|] + C

\int \frac{\sqrt{x^{2} - 25}}{x} d x

\int \frac{θ^{3} d θ}{\sqrt{9 - θ^{2}}} d θ

- \frac{1}{3} \sqrt{9 - θ^{2}} (18 + θ^{2}) + C

\int \frac{d x}{\sqrt{x^{6} - x^{2}}}

\int \sqrt{x^{6} - x^{8}} d x

\frac{(−1 + x^{2}) (2 + 3 x^{2}) \sqrt{x^{6} - x^{8}}}{15 x^{3}} + C

\int \frac{d x}{{(1 + x^{2})}^{3 / 2}}

\int \frac{d x}{{(x^{2} - 9)}^{3 / 2}}

- \frac{x}{9 \sqrt{−9 + x^{2}}} + C

\int \frac{\sqrt{1 + x^{2}} d x}{x}

\int \frac{x^{2} d x}{\sqrt{x^{2} - 1}}

\frac{1}{2} (ln \| x + \sqrt{x^{2} - 1} \| + x \sqrt{x^{2} - 1}) + C

\int \frac{x^{2} d x}{x^{2} + 4}

\int \frac{d x}{x^{2} \sqrt{x^{2} + 1}}

- \frac{\sqrt{1 + x^{2}}}{x} + C

\int \frac{x^{2} d x}{\sqrt{1 + x^{2}}}

\int_{−1}^{1} {(1 - x^{2})}^{3 / 2} d x

\frac{1}{8} (x (5 - 2 x^{2}) \sqrt{1 - x^{2}} + 3 arcsin x) + C

In the following exercises, use the substitutions $x = sinh θ, cosh θ,$

or $tanh θ .$

Express the final answers in terms of the variable x.

\int \frac{d x}{\sqrt{x^{2} - 1}}

\int \frac{d x}{x \sqrt{1 - x^{2}}}

ln x - ln \| 1 + \sqrt{1 - x^{2}} \| + C

\int \sqrt{x^{2} - 1} d x

\int \frac{\sqrt{x^{2} - 1}}{x^{2}} d x

- \frac{\sqrt{−1 + x^{2}}}{x} + ln \| x + \sqrt{−1 + x^{2}} \| + C

\int \frac{d x}{1 - x^{2}}

\int \frac{\sqrt{1 + x^{2}}}{x^{2}} d x

- \frac{\sqrt{1 + x^{2}}}{x} + arcsinh x + C

Use the technique of completing the square to evaluate the following integrals.

\int \frac{1}{x^{2} - 6 x} d x

\int \frac{1}{x^{2} + 2 x + 1} d x

- \frac{1}{1 + x} + C

\int \frac{1}{\sqrt{− x^{2} + 2 x + 8}} d x

\int \frac{1}{\sqrt{− x^{2} + 10 x}} d x

\frac{2 \sqrt{−10 + x} \sqrt{x} ln \| \sqrt{−10 + x} + \sqrt{x} \|}{\sqrt{(10 - x) x}} + C

\int \frac{1}{\sqrt{x^{2} + 4 x - 12}} d x

Evaluate the integral without using calculus: $\int_{−3}^{3} \sqrt{9 - x^{2}} d x .$

\frac{9 π}{2};

area of a semicircle with radius 3

Find the area enclosed by the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1 .$

Evaluate the integral $\int \frac{d x}{\sqrt{1 - x^{2}}}$

using two different substitutions. First, let $x = cos θ$

and evaluate using trigonometric substitution. Second, let $x = sin θ$

and use trigonometric substitution. Are the answers the same?

arcsin (x) + C

is the common answer.

Evaluate the integral $\int \frac{d x}{x \sqrt{x^{2} - 1}}$

using the substitution $x = sec θ .$

Next, evaluate the same integral using the substitution $x = csc θ .$

Show that the results are equivalent.

Evaluate the integral $\int \frac{x}{x^{2} + 1} d x$

using the form $\int \frac{1}{u} d u .$

Next, evaluate the same integral using $x = tan θ .$

Are the results the same?

\frac{1}{2} ln (1 + x^{2}) + C

is the result using either method.

State the method of integration you would use to evaluate the integral $\int x \sqrt{x^{2} + 1} d x .$

Why did you choose this method?

State the method of integration you would use to evaluate the integral $\int x^{2} \sqrt{x^{2} - 1} d x .$

Why did you choose this method?

Use trigonometric substitution. Let $x = sec (θ) .$

Evaluate $\int_{−1}^{1} \frac{x d x}{x^{2} + 1}$

Find the length of the arc of the curve over the specified interval: $y = ln x, [1, 5] .$

Round the answer to three decimal places.

4.367

Find the surface area of the solid generated by revolving the region bounded by the graphs of $y = x^{2}, y = 0, x = 0, and x = \sqrt{2}$

about the x-axis. (Round the answer to three decimal places).

The region bounded by the graph of $f (x) = \frac{1}{1 + x^{2}}$

and the x-axis between $x = 0$

and $x = 1$

is revolved about the x-axis. Find the volume of the solid that is generated.

\frac{π^{2}}{8} + \frac{π}{4}

Solve the initial-value problem for y as a function of x.

(x^{2} + 36) \frac{d y}{d x} = 1, y (6) = 0

(64 - x^{2}) \frac{d y}{d x} = 1, y (0) = 3

y = \frac{1}{16} ln \| \frac{x + 8}{x - 8} \| + 3

Find the area bounded by $y = \frac{2}{\sqrt{64 - 4 x^{2}}}, x = 0, y = 0, and x = 2 .$

An oil storage tank can be described as the volume generated by revolving the area bounded by $y = \frac{16}{\sqrt{64 + x^{2}}}, x = 0, y = 0, x = 2$

about the x-axis. Find the volume of the tank (in cubic meters).

24.6 m³

During each cycle, the velocity v (in feet per second) of a robotic welding device is given by $v = 2 t - \frac{14}{4 + t^{2}},$

where t is time in seconds. Find the expression for the displacement s (in feet) as a function of t if $s = 0$

when $t = 0 .$

Find the length of the curve $y = \sqrt{16 - x^{2}}$

between $x = 0$

and $x = 2 .$

\frac{2 π}{3}

Trigonometric Substitution

Integrals Involving a2−x2

Integrating Expressions Involving a2+x2

Integrating Expressions Involving x2−a2

Key Concepts

Glossary

Integrals Involving $\sqrt{a^{2} - x^{2}}$

Integrating Expressions Involving $\sqrt{a^{2} + x^{2}}$

Integrating Expressions Involving $\sqrt{x^{2} - a^{2}}$