Math · Calculus II · Worked example

Integrate √(9 − x²) by trig substitution

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

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

Choose the substitution

$9 - x^{2}$ has the form $a^{2} - x^{2}$ with $a = 3$ , and there is no factor of $x$ to pair with $d x$ , so an ordinary substitution fails. Let $x = 3 sin θ$ , with $θ$ between $- π / 2$ and $π / 2$ .

x = 3 sin θ

Find $d x$

Differentiate the substitution. The factor $3 cos θ$ goes into the integral with $d θ$ .

d x = 3 cos θ d θ

Simplify the root

$9 - 9 {sin}^{2} θ = 9 {cos}^{2} θ$ , and $cos θ$ is not negative for $θ$ between $- π / 2$ and $π / 2$ , so the root is $3 cos θ$ .

\sqrt{9 - 9 {sin}^{2} θ} = 3 cos θ

Integrate in $θ$

The integrand is $(3 cos θ) (3 cos θ)$ $= 9 {cos}^{2} θ$ . Since ${cos}^{2} θ = (1 + cos 2 θ) / 2$ , that is $(9 / 2) (1 + cos 2 θ)$ , which has a known antiderivative.

\int \frac{9}{2} (1 + cos 2 θ) d θ = \frac{9}{2} θ + \frac{9}{4} sin 2 θ + C

\begin{array}{c} \int \frac{9}{2} (1 + cos 2 θ) d θ \\ = \frac{9}{2} θ + \frac{9}{4} sin 2 θ + C \end{array}

Return to $x$

$θ = arcsin (x / 3)$ . For $sin 2 θ$ , use $sin 2 θ = 2 sin θ cos θ$ and read the triangle: $sin θ = x / 3$ and $cos θ = \sqrt{9 - x^{2}} / 3$ . Then $(9 / 4) sin 2 θ$ $= (9 / 4) (2) (x / 3) (\sqrt{9 - x^{2}} / 3)$ $= x \sqrt{9 - x^{2}} / 2$ .

\frac{9}{2} arcsin \frac{x}{3} + \frac{x \sqrt{9 - x^{2}}}{2} + C

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

Check a value with an area

Differentiating the answer returns $\sqrt{9 - x^{2}}$ , which checks it for every $x$ . A quicker check of one value: from $x = 0$ to $x = 3$ the antiderivative gives $(9 / 2) (π / 2) = 9 π / 4$ . The graph of $y = \sqrt{9 - x^{2}}$ on that interval is a quarter of a circle of radius $3$ , and a quarter of $π \cdot 3^{2}$ is also $9 π / 4$ .

Result

$\int \sqrt{9 - x^{2}} d x$ $= (9 / 2) arcsin (x / 3)$ $+ (x / 2) \sqrt{9 - x^{2}} + C$ , for $x$ from $- 3$ to $3$ . From $0$ to $3$ it gives $9 π / 4$ , the quarter circle.

Your turn

1. One like the example

Find $\int d x / \sqrt{25 - x^{2}}$ .

Show the answer and explanation

$arcsin (x / 5) + C$ , for $x$ between $- 5$ and $5$ .

Let $x = 5 sin θ$ , so $d x = 5 cos θ d θ$ and $\sqrt{25 - x^{2}} = 5 cos θ$ . The integrand becomes $(5 cos θ) / (5 cos θ) = 1$ , so the integral is $θ + C$ , and $θ = arcsin (x / 5)$ . Chalk checks every row, the $arcsin$ answer included.

\begin{array}{l} \int \frac{1}{\sqrt{25 - x^{2}}} d x \\ x = 5 sin θ \\ d x = 5 cos θ d θ \\ \int 1 d θ = θ + C \\ arcsin \frac{x}{5} + C \end{array}

2. A variation

Find $\int x^{2} / \sqrt{9 - x^{2}} d x$ . The root has moved to the denominator, with $x^{2}$ above it.

Show the answer and explanation

$(9 / 2) arcsin (x / 3)$ $- (x / 2) \sqrt{9 - x^{2}} + C$ .

With $x = 3 sin θ$ , $d x = 3 cos θ d θ$ and the root is $3 cos θ$ , so the integrand becomes $9 {sin}^{2} θ \cdot 3 cos θ / (3 cos θ)$ $= 9 {sin}^{2} θ$ . Since ${sin}^{2} θ = (1 - cos 2 θ) / 2$ , the θ-integral is $(9 / 2) θ - (9 / 4) sin 2 θ$ . The triangle gives $(9 / 4) sin 2 θ$ $= x \sqrt{9 - x^{2}} / 2$ , as in the worked example.

\begin{array}{l} \int \frac{x^{2}}{\sqrt{9 - x^{2}}} d x \\ x = 3 sin θ \\ d x = 3 cos θ d θ \\ \int \frac{9}{2} (1 - cos 2 θ) d θ = \frac{9}{2} θ - \frac{9}{4} sin 2 θ + C \\ \frac{9}{2} arcsin \frac{x}{3} - \frac{x \sqrt{9 - x^{2}}}{2} + C \end{array}

\begin{array}{c} \int \frac{x^{2}}{\sqrt{9 - x^{2}}} d x \\ x = 3 sin θ \\ d x = 3 cos θ d θ \\ \int \frac{9}{2} (1 - cos 2 θ) d θ \\ = \frac{9}{2} θ - \frac{9}{4} sin 2 θ + C \\ = \frac{9}{2} arcsin \frac{x}{3} - \frac{x \sqrt{9 - x^{2}}}{2} + C \end{array}

3. Somewhere new

Find the area inside the ellipse $x^{2} / 16 + y^{2} / 9 = 1$ .

Show the answer and explanation

$12 π \approx 37.7$ square units.

The top half is $y = (3 / 4) \sqrt{16 - x^{2}}$ , so the area is four times the integral of $(3 / 4) \sqrt{16 - x^{2}}$ from $0$ to $4$ , which is $3$ times the integral of $\sqrt{16 - x^{2}}$ . With $x = 4 sin θ$ that integral is $\int 16 {cos}^{2} θ d θ$ from $0$ to $π / 2$ , or $4 π$ , a quarter of a circle of radius $4$ . The area is $3 \cdot 4 π = 12 π$ , which matches πab with $a = 4$ and $b = 3$ .

\begin{array}{l} \int_{0}^{4} \sqrt{16 - x^{2}} d x \\ x = 4 sin θ \\ d x = 4 cos θ d θ \\ \int_{0}^{π / 2} 8 (1 + cos 2 θ) d θ \\ {[8 θ + 4 sin 2 θ]}_{0}^{π / 2} \\ 4 π \\ 3 \cdot 4 π = 12 π \\ 12 π \approx 37.7 \end{array}

Keep exploring

Open the substitution in Math and change the differential to $d x = 3 sin θ d θ$ . The row turns to ✗ with “Doesn’t match dx = x′(θ) dθ.” Add the final answer as a row and it reads ✓, because the answer is checked against the integral in row 1, not against the differential.

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 in Math Check the θ-integral in Math See the quarter circle in Graph 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

Return to xx

Check a value with an area

Result

Your turn

1. One like the example

arcsin(x⁢/5)+Carcsin(x/5)+C, for xx between −5−5 and 55.

2. A variation

(9⁢/2)arcsin(x⁢/3)(9/2)arcsin(x/3) −(x⁢/2)9−x2+C−(x/2)√9−x2​+C.

3. Somewhere new

12⁢π≈37.712π≈37.7 square units.

Keep exploring

Study it, then try it

Find $d x$

Integrate in $θ$

Return to $x$

$arcsin (x / 5) + C$ , for $x$ between $- 5$ and $5$ .

$(9 / 2) arcsin (x / 3)$ $- (x / 2) \sqrt{9 - x^{2}} + C$ .

$12 π \approx 37.7$ square units.