3.3 Trigonometric substitution (Page 3/5)

Calculus volume 2 Page 3 / 5

Problem-solving strategy: integrating expressions involving $\sqrt{a^{2} + x^{2}}$

Check to see whether the integral can be evaluated easily by using another method. In some cases, it is more convenient to use an alternative method.
Substitute $x = a tan θ$ and $d x = a {sec}^{2} θ d θ .$ This substitution yields
$\sqrt{a^{2} + x^{2}} = \sqrt{a^{2} + {(a tan θ)}^{2}} = \sqrt{a^{2} (1 + {tan}^{2} θ)} = \sqrt{a^{2} {sec}^{2} θ} = | a sec θ | = a sec θ .$ (Since $- \frac{π}{2} < θ < \frac{π}{2}$ and $sec θ > 0$ over this interval, $| a sec θ | = a sec θ .)$
Simplify the expression.
Evaluate the integral using techniques from the section on trigonometric integrals.
Use the reference triangle from [link] to rewrite the result in terms of $x .$ You may also need to use some trigonometric identities and the relationship $θ = {tan}^{−1} (\frac{x}{a}) .$ ( Note : The reference triangle is based on the assumption that $x > 0;$ however, the trigonometric ratios produced from the reference triangle are the same as the ratios for which $x \leq 0.)$

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 (a^2+x^2), the vertical leg is labeled x, and the horizontal leg is labeled a. To the left of the triangle is the equation tan(theta) = x/a. — A reference triangle can be constructed to express the trigonometric functions evaluated at $θ$ in terms of $x .$

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.

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.

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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}

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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}

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Rewrite $\int^{} x^{3} \sqrt{x^{2} + 4} d x$ by using a substitution involving $tan θ .$

$\int^{} 32 {tan}^{3} θ {sec}^{3} θ d θ$

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Integrating expressions involving $\sqrt{x^{2} - a^{2}}$

The domain of the expression $\sqrt{x^{2} - a^{2}}$ is $(− \infty, − a] \cup [a, + \infty) .$ Thus, either $x < − a$ or $x > a .$ Hence, $\frac{x}{a} \leq - 1$ or $\frac{x}{a} \geq 1 .$ Since these intervals correspond to the range of $sec θ$ on the set $[0, \frac{π}{2}) \cup (\frac{π}{2}, π],$ it makes sense to use the substitution $sec θ = \frac{x}{a}$ or, equivalently, $x = a sec θ,$ where $0 \leq θ < \frac{π}{2}$ or $\frac{π}{2} < θ \leq π .$ The corresponding substitution for $d x$ is $d x = a sec θ tan θ d θ .$ The procedure for using this substitution is outlined in the following problem-solving strategy.

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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3.3 Trigonometric substitution (Page 3/5)

Problem-solving strategy: integrating expressions involving a 2 + x 2

Integrating an expression involving a 2 + x 2

Evaluating ∫ d x 1 + x 2 Using a different substitution

Finding an arc length

Integrating expressions involving x 2 − a 2

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Problem-solving strategy: integrating expressions involving $\sqrt{a^{2} + x^{2}}$

Integrating an expression involving $\sqrt{a^{2} + x^{2}}$

Evaluating $\int \frac{d x}{\sqrt{1 + x^{2}}}$ Using a different substitution

Integrating expressions involving $\sqrt{x^{2} - a^{2}}$