Math · Calculus II · Worked example

Use a secant substitution and a triangle

Find $\int d x / (x^{2} \sqrt{x^{2} - 9})$ for $x > 3$ .

\int \frac{d x}{x^{2} \sqrt{x^{2} - 9}}

Choose the substitution

$x^{2} - 9$ has the form $x^{2} - a^{2}$ with $a = 3$ , so the secant fits. Let $x = 3 sec θ$ , with $θ$ between $0$ and $π / 2$ because $x > 3$ .

x = 3 sec θ

Find $d x$

The derivative of $sec θ$ is $sec θ tan θ$ , so $d x = 3 sec θ tan θ d θ$ . This factor goes into the integral in place of $d x$ .

d x = 3 sec θ tan θ d θ

Simplify the root

$x^{2} - 9$ $= 9 {sec}^{2} θ - 9$ $= 9 {tan}^{2} θ$ , and $tan θ$ is positive on this range, so the root is $3 tan θ$ .

\sqrt{9 {sec}^{2} θ - 9} = 3 tan θ

Integrate in $θ$

The integrand is $3 sec θ tan θ$ over $(9 {sec}^{2} θ) (3 tan θ)$ . The tangents cancel and one secant stays in the denominator, which leaves $(1 / 9) cos θ$ .

\int \frac{1}{9} cos θ d θ = \frac{1}{9} sin θ + C

Read $sin θ$ from the triangle

$sec θ = x / 3$ means the hypotenuse is $x$ and the side adjacent to $θ$ is $3$ , so the opposite side is $\sqrt{x^{2} - 9}$ . Sine is opposite over hypotenuse.

sin θ = \frac{\sqrt{x^{2} - 9}}{x}

The triangle for $x$ $= 3 sec θ$
Side	Length
Hypotenuse	$x$
Adjacent to $θ$	$3$
Opposite $θ$	$\sqrt{x^{2} - 9}$

Write the answer and check it

Substitute for $sin θ$ . To check, differentiate $\sqrt{x^{2} - 9} / (9 x)$ with the quotient rule: the result simplifies to $(x^{2} - (x^{2} - 9)) / (9 x^{2} \sqrt{x^{2} - 9})$ $= 1 / (x^{2} \sqrt{x^{2} - 9})$ , the integrand.

\frac{\sqrt{x^{2} - 9}}{9 x} + C

Result

$\int d x / (x^{2} \sqrt{x^{2} - 9})$ $= \sqrt{x^{2} - 9} / (9 x) + C$ for $x > 3$ . The answer starts at $0$ when $x = 3$ and creeps up toward $1 / 9$ as $x$ grows.

Your turn

1. One like the example

Find $\int d x / (x \sqrt{x^{2} - 4})$ for $x > 2$ .

Show the answer and explanation

$(1 / 2) arccos (2 / x) + C$ , also written $(1 / 2) arcsec (x / 2) + C$ .

Let $x = 2 sec θ$ , so $d x = 2 sec θ tan θ d θ$ and $\sqrt{x^{2} - 4} = 2 tan θ$ . The integrand becomes $(2 sec θ tan θ) / ((2 sec θ) (2 tan θ))$ $= 1 / 2$ , so the integral is $θ / 2 + C$ . Since $sec θ = x / 2$ , $cos θ = 2 / x$ and $θ = arccos (2 / x)$ . Chalk checks the θ-integral and leaves the $arccos$ answer at “?” here. Written straight under $\int d x / (x \sqrt{x^{2} - 4})$ , it reads ✓.

\begin{array}{l} \int \frac{1}{x \sqrt{x^{2} - 4}} d x \\ x = 2 sec θ \\ d x = 2 sec θ tan θ d θ \\ \int \frac{1}{2} d θ = \frac{θ}{2} + C \\ \frac{1}{2} arccos \frac{2}{x} + C \end{array}

2. A variation

Find $\int d x / {(x^{2} - 4)}^{3 / 2}$ for $x > 2$ . The root is now raised to a power.

Show the answer and explanation

$- x / (4 \sqrt{x^{2} - 4}) + C$ .

With $x = 2 sec θ$ , ${(x^{2} - 4)}^{3 / 2}$ $= {(2 tan θ)}^{3}$ $= 8 {tan}^{3} θ$ , and $d x = 2 sec θ tan θ d θ$ . The integrand becomes $sec θ / (4 {tan}^{2} θ)$ $= (1 / 4) csc θ cot θ$ , whose antiderivative is $- (1 / 4) csc θ$ . The triangle gives $csc θ = x / \sqrt{x^{2} - 4}$ . Chalk checks the θ-integral and this answer in $x$ .

\begin{array}{l} \int \frac{1}{{(x^{2} - 4)}^{3 / 2}} d x \\ x = 2 sec θ \\ d x = 2 sec θ tan θ d θ \\ \int \frac{1}{4} csc θ cot θ d θ = - \frac{1}{4} csc θ + C \\ - \frac{x}{4 \sqrt{x^{2} - 4}} + C \end{array}

3. Somewhere new

Evaluate $\int d x / (x^{2} \sqrt{x^{2} - 9})$ from $x = 3$ to $x = 6$ . Change the limits instead of returning to $x$ .

Show the answer and explanation

$\sqrt{3} / 18 \approx 0.0962$ .

With $x = 3 sec θ$ , $x = 3$ gives $sec θ = 1$ , so $θ = 0$ , and $x = 6$ gives $sec θ = 2$ , so $θ = π / 3$ . The θ-integral of $(1 / 9) cos θ$ from $0$ to $π / 3$ is $(1 / 9) sin (π / 3)$ $= (1 / 9) (\sqrt{3} / 2)$ $= \sqrt{3} / 18$ . The worked example’s answer agrees, $\sqrt{27} / 54 - 0 = \sqrt{3} / 18$ .

\begin{array}{l} \int_{3}^{6} \frac{1}{x^{2} \sqrt{x^{2} - 9}} d x \\ x = 3 sec θ \\ d x = 3 sec θ tan θ d θ \\ \int_{0}^{π / 3} \frac{1}{9} cos θ d θ \\ \frac{1}{9} sin \frac{π}{3} = \frac{\sqrt{3}}{18} \\ \frac{\sqrt{3}}{18} \approx 0.0962 \end{array}

Keep exploring

Check it in Math, then change the answer to $1 / (3 x) + C$ , the answer you get by reading $sin θ$ as $3 / x$ , which is $cos θ$ . The row turns to ✗: it isn’t an antiderivative of the integrand.

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Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Strang, Calculus (MIT OpenCourseWare) — §7.3 Trigonometric Substitutions

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 6 key terms Check it by differentiating in Math Check it in Math Check the substitution in Math Open worked example on a board Pythagorean identity in Math Reference

Your existing work stays on this device. Examples open as editable copies.

Choose the substitution

Find d⁢xdx

Simplify the root

Integrate in θθ

Read sinθsinθ from the triangle

Write the answer and check it

Result

Your turn

1. One like the example

(1⁢/2)arccos(2⁢/⁢x)+C(1/2)arccos(2/x)+C, also written (1⁢/2)⁢arcsec⁢(x⁢/2)+C(1/2)arcsec(x/2)+C.

2. A variation

−x⁢/⁢(4x2−4)+C−x/(4√x2−4​)+C.

3. Somewhere new

3/18≈0.0962√3/18≈0.0962.

Keep exploring

Study it, then try it

Find $d x$

Integrate in $θ$

Read $sin θ$ from the triangle

$(1 / 2) arccos (2 / x) + C$ , also written $(1 / 2) arcsec (x / 2) + C$ .

$- x / (4 \sqrt{x^{2} - 4}) + C$ .

$\sqrt{3} / 18 \approx 0.0962$ .