Math · Calculus II · Worked example

Change the limits in a tangent substitution

Evaluate $\int_{0}^{2} d x / {(x^{2} + 4)}^{3 / 2}$ .

\int_{0}^{2} \frac{d x}{{(x^{2} + 4)}^{3 / 2}}

Choose the substitution

$x^{2} + 4$ has the form $x^{2} + a^{2}$ with $a = 2$ . Let $x = 2 tan θ$ , with $θ$ between $- π / 2$ and $π / 2$ . Then $d x = 2 {sec}^{2} θ d θ$ .

\begin{array}{l} x = 2 tan θ \\ d x = 2 {sec}^{2} θ d θ \end{array}

Simplify the denominator

$x^{2} + 4$ $= 4 {tan}^{2} θ + 4$ $= 4 {sec}^{2} θ$ . Its $3 / 2$ power is ${(2 sec θ)}^{3} = 8 {sec}^{3} θ$ , because $sec θ$ is positive on this range.

{(4 {tan}^{2} θ + 4)}^{3 / 2} = 8 {sec}^{3} θ

Convert the limits

When $x = 0$ , $tan θ = 0$ , so $θ = 0$ . When $x = 2$ , $tan θ = 1$ , so $θ = π / 4$ .

x = 0 \to θ = 0, x = 2 \to θ = \frac{π}{4}

Integrate in $θ$

The integrand is $2 {sec}^{2} θ / (8 {sec}^{3} θ)$ $= (1 / 4) cos θ$ , and an antiderivative of $cos θ$ is $sin θ$ .

\int_{0}^{π / 4} \frac{1}{4} cos θ d θ = \frac{1}{4} sin \frac{π}{4} = \frac{\sqrt{2}}{8}

Stay in $θ$

The limits are already values of $θ$ , so no triangle is needed. As a check, returning to $x$ gives the antiderivative $x / (4 \sqrt{x^{2} + 4})$ , and at $x = 2$ that is $2 / (4 \sqrt{8}) = \sqrt{2} / 8$ , the same value.

Result

$\int_{0}^{2} d x / {(x^{2} + 4)}^{3 / 2}$ $= \sqrt{2} / 8$ $\approx 0.177$ .

Your turn

1. One like the example

Evaluate $\int_{0}^{3} d x / (x^{2} + 9)$ .

Show the answer and explanation

$π / 12 \approx 0.262$ .

Let $x = 3 tan θ$ , so $d x = 3 {sec}^{2} θ d θ$ and $x^{2} + 9 = 9 {sec}^{2} θ$ . The integrand becomes $3 {sec}^{2} θ / (9 {sec}^{2} θ) = 1 / 3$ , and the limits become $θ = 0$ and $θ = π / 4$ , because $tan (π / 4) = 1$ . The integral is $(1 / 3) (π / 4) = π / 12$ . In a Math box the row that moves to $θ$ reads ✓ “Row 1, written in θ: x from 0 to 3 is θ from 0 to π/4.”, and the value after it reads ✓.

\begin{array}{l} \int_{0}^{3} \frac{1}{x^{2} + 9} d x \\ x = 3 tan θ \\ d x = 3 {sec}^{2} θ d θ \\ \int_{0}^{π / 4} \frac{1}{3} d θ \\ \frac{π}{12} \end{array}

2. A variation

Evaluate $\int_{0}^{1} d x / \sqrt{x^{2} + 1}$ . The θ-integral that comes out is a secant.

Show the answer and explanation

$ln (1 + \sqrt{2}) \approx 0.881$ .

With $x = tan θ$ , $d x = {sec}^{2} θ d θ$ and $\sqrt{x^{2} + 1} = sec θ$ , so the integrand becomes $sec θ$ . The limits are $θ = 0$ and $θ = π / 4$ . $ln ∣ sec θ + tan θ ∣$ is $ln (\sqrt{2} + 1)$ at $π / 4$ and $ln 1 = 0$ at $0$ . The row with the secant integral and its value reads ✓ “Row 1, written in θ: x from 0 to 1 is θ from 0 to π/4.”, and the decimal after it reads ✓.

\begin{array}{l} \int_{0}^{1} \frac{1}{\sqrt{x^{2} + 1}} d x \\ x = tan θ \\ d x = {sec}^{2} θ d θ \\ \int_{0}^{π / 4} sec θ d θ = ln (\sqrt{2} + 1) \\ ln (\sqrt{2} + 1) \approx 0.881 \end{array}

3. Somewhere new

A straight rod of charge runs from $x = - 3$ to $x = 3$ . Its pull on a point $4$ units from its middle is proportional to $\int d x / {(x^{2} + 16)}^{3 / 2}$ from $- 3$ to $3$ . Evaluate the integral.

Show the answer and explanation

$3 / 40 = 0.075$ .

Let $x = 4 tan θ$ , so $d x = 4 {sec}^{2} θ d θ$ and ${(x^{2} + 16)}^{3 / 2}$ $= 64 {sec}^{3} θ$ . The integrand becomes $(1 / 16) cos θ$ , with antiderivative $(1 / 16) sin θ$ . At $x = 3$ , $tan θ = 3 / 4$ , and a 3-4-5 triangle gives $sin θ = 3 / 5$ . At $x = - 3$ , $sin θ = - 3 / 5$ . The value is $(1 / 16) (3 / 5 + 3 / 5) = 3 / 40$ .

\begin{array}{l} \int_{- 3}^{3} \frac{1}{{(x^{2} + 16)}^{3 / 2}} d x \\ x = 4 tan θ \\ d x = 4 {sec}^{2} θ d θ \\ \frac{1}{16} (\frac{3}{5} + \frac{3}{5}) = \frac{3}{40} \end{array}

Keep exploring

Slopes, sums & signed area integrates the original integrand in $x$ and gets about $0.1768$ , which is $\sqrt{2} / 8$ . In the Math box, change $\sqrt{2} / 8$ to $\sqrt{2} / 2$ , the value with the factor $1 / 4$ forgotten, and the row turns to ✗.

Return to the concept →

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 the substitution and the θ-integral in Math Check the θ-integral in Math Integrate in x in Slopes, sums & signed area 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

Simplify the denominator

Convert the limits

Integrate in θθ

Stay in θθ

Result

Your turn

1. One like the example

π⁢/12≈0.262π/12≈0.262.

2. A variation

ln(1+2)≈0.881ln(1+√2)≈0.881.

3. Somewhere new

3⁢/40=0.0753/40=0.075.

Keep exploring

Study it, then try it

Integrate in $θ$

Stay in $θ$

$π / 12 \approx 0.262$ .

$ln (1 + \sqrt{2}) \approx 0.881$ .

$3 / 40 = 0.075$ .