Math · Calculus II · Concept

Trigonometric substitution

Trigonometric substitution handles integrals that contain $\sqrt{a^{2} - x^{2}}$ , $\sqrt{a^{2} + x^{2}}$ or $\sqrt{x^{2} - a^{2}}$ . Replace $x$ with $a sin θ$ , $a tan θ$ or $a sec θ$ , and an identity such as $1 - {sin}^{2} θ = {cos}^{2} θ$ turns the expression under the root into a perfect square. Integrate in $θ$ , then return to $x$ with a right triangle, or change the limits of a definite integral and stay in $θ$ .

Before this

Integration by substitution (u-substitution). Substitution renames an inside expression $u$ and replaces $g^{'} (x) d x$ with $d u$ , so the differential changes along with the variable.
Trigonometric identities and equations. The Pythagorean identity ${sin}^{2} θ + {cos}^{2} θ = 1$ , divided by ${cos}^{2} θ$ , also gives ${tan}^{2} θ + 1 = {sec}^{2} θ$ .
Composite and inverse functions. arcsin, arctan and arccos undo sine, tangent and cosine on a restricted range, so $arcsin (sin θ) = θ$ only for $θ$ between $- π / 2$ and $π / 2$ .

A concrete case

A quarter circle, measured in angles

The curve $y = \sqrt{9 - x^{2}}$ from $x = 0$ to $x = 3$ is a quarter of a circle of radius $3$ . Geometry gives its area as a quarter of $π \cdot 3^{2}$ , which is $9 π / 4 \approx 7.069$ . An integral should agree, but the substitution $u = 9 - x^{2}$ needs a factor of $x$ beside $d x$ , and there is none.

Describe each point of the circle by an angle instead. Put $x = 3 sin θ$ . The height $\sqrt{9 - x^{2}}$ becomes $\sqrt{9 - 9 {sin}^{2} θ} = 3 cos θ$ , with no root left. At $θ = π / 6$ , for example, $x = 1.5$ and the height is $\sqrt{6.75} \approx 2.598$ , which is $3 cos (π / 6)$ .

The width of a strip changes too. As $θ$ moves by a small $d θ$ , $x$ moves by about $3 cos θ d θ$ , so equal steps in $θ$ give narrower strips in $x$ near $x = 3$ , where $θ$ is close to $π / 2$ . The area is the integral of the height $3 cos θ$ times the width $3 cos θ d θ$ , from $θ = 0$ to $θ = π / 2$ . That integral of $9 {cos}^{2} θ$ is $9 π / 4$ , the quarter circle exactly.

Forgetting the width gives the integral of $3 cos θ$ alone, which is $3$ , far from $7.069$ . The root disappears only when $d x$ changes along with it.

\int_{0}^{3} \sqrt{9 - x^{2}} d x = \int_{0}^{π / 2} 9 {cos}^{2} θ d θ = \frac{9 π}{4}

\begin{aligned} \int_{0}^{3} \sqrt{9 - x^{2}} d x \\ = \int_{0}^{π / 2} 9 {cos}^{2} θ d θ \\ = \frac{9 π}{4} \end{aligned}

The rule

The three substitutions

Trigonometric substitution replaces $x$ with a times a trigonometric function of a new variable $θ$ . The function is chosen so that a Pythagorean identity turns the expression under the root into one square. Here $a$ is a positive constant read from the integral, as in $9 - x^{2} = 3^{2} - x^{2}$ .

$θ$ is kept in a range where the substitution can be undone and the root comes out without an absolute value. The range is $- π / 2 \leq θ \leq π / 2$ for sine, $- π / 2 < θ < π / 2$ for tangent, and $0 \leq θ < π / 2$ for secant when $x \geq a$ .

\sqrt{a^{2} - a^{2} {sin}^{2} θ} = a cos θ

The three trigonometric substitutions
Integral contains	Substitute	$d x$	The root becomes
$\sqrt{a^{2} - x^{2}}$	$x$ $= a sin θ$	$a cos θ d θ$	$a cos θ$
$\sqrt{a^{2} + x^{2}}$	$x$ $= a tan θ$	$a {sec}^{2} θ d θ$	$a sec θ$
$\sqrt{x^{2} - a^{2}}$	$x$ $= a sec θ$	$a sec θ tan θ d θ$	$a tan θ$

Replace $d x$ as well

The substitution changes the variable of integration, so $d x$ changes with it. Differentiate $x = a sin θ$ to get $d x = a cos θ d θ$ . Swapping the root for $a cos θ$ and writing $d θ$ in place of $d x$ gives an integral with a different value, as the quarter circle showed.

x = a sin θ \Rightarrow d x = a cos θ d θ

Why it works

Each Pythagorean identity rewrites a sum or difference of squares as a single square. The identities are $1 - {sin}^{2} θ = {cos}^{2} θ$ , $1 + {tan}^{2} θ = {sec}^{2} θ$ and ${sec}^{2} θ - 1 = {tan}^{2} θ$ . Multiplied through by $a^{2}$ , they match the forms $a^{2} - x^{2}$ , $a^{2} + x^{2}$ and $x^{2} - a^{2}$ , and the square root of a single square is that quantity up to its sign.

The change of variable is substitution run in the other direction. In u-substitution an inside expression gets a new name. Here $x$ itself is written as a function of $θ$ , $x = g (θ)$ , and the chain rule gives $d x = g^{'} (θ) d θ$ . That direction needs $g$ to be one-to-one on the chosen range, so that each $x$ comes from exactly one $θ$ and the answer can return to $x$ . On each range the function the root becomes is not negative: $cos θ$ for the sine substitution, $sec θ$ for the tangent and $tan θ$ for the secant. That is why the root loses its absolute value.

\int f (x) d x = \int f (g (θ)) g^{'} (θ) d θ

\begin{array}{c} \int f (x) d x \\ = \int f (g (θ)) g^{'} (θ) d θ \end{array}

The method

The method in seven steps

The table works through $\int d x / (x^{2} \sqrt{4 - x^{2}})$ for $x$ between $0$ and $2$ .

\int \frac{d x}{x^{2} \sqrt{4 - x^{2}}} = \int \frac{1}{4} {csc}^{2} θ d θ = - \frac{\sqrt{4 - x^{2}}}{4 x} + C

\begin{aligned} \int \frac{d x}{x^{2} \sqrt{4 - x^{2}}} \\ = \int \frac{1}{4} {csc}^{2} θ d θ \\ = - \frac{\sqrt{4 - x^{2}}}{4 x} + C \end{aligned}

Trigonometric substitution on the integral of $1 / (x^{2} \sqrt{4 - x^{2}})$
Step	What to do	For $\int d x / (x^{2} \sqrt{4 - x^{2}})$
1	Match the form and choose $x$	$4 - x^{2}$ $= 2^{2} - x^{2}$ , so $x$ $= 2 sin θ$
2	Find $d x$	$d x$ $= 2 cos θ d θ$
3	Simplify the root	$\sqrt{4 - 4 {sin}^{2} θ}$ $= 2 cos θ$
4	Rewrite the integral in $θ$	$\int 2 cos θ d θ / (4 {sin}^{2} θ \cdot 2 cos θ)$ $= \int \frac{1}{4} {csc}^{2} θ d θ$
5	Integrate in $θ$	$- \frac{1}{4} cot θ + C$
6	Return to $x$ with the triangle	$cot θ$ $= \sqrt{4 - x^{2}} / x$ , so $- \sqrt{4 - x^{2}} / (4 x) + C$
7	Check by differentiating	The derivative is $1 / (x^{2} \sqrt{4 - x^{2}})$

Return to $x$ with a triangle

An indefinite integral in $x$ needs an answer in $x$ . Draw a right triangle that shows the substitution. $x = a sin θ$ says $sin θ = x / a$ , so the side opposite $θ$ is $x$ and the hypotenuse is $a$ . The Pythagorean theorem gives the third side, and every other trigonometric function of $θ$ can be read from the triangle. $θ$ itself becomes $arcsin (x / a)$ , $arctan (x / a)$ or $arccos (a / x)$ for the three substitutions.

Sides of the triangle for each substitution
Substitution	Opposite $θ$	Adjacent to $θ$	Hypotenuse
$x$ $= a sin θ$	$x$	$\sqrt{a^{2} - x^{2}}$	$a$
$x$ $= a tan θ$	$x$	$a$	$\sqrt{a^{2} + x^{2}}$
$x$ $= a sec θ$	$\sqrt{x^{2} - a^{2}}$	$a$	$x$

Definite integrals: change the limits

For a definite integral, convert each limit to a value of $θ$ and evaluate there. With $x = 2 tan θ$ , the limit $x = 2$ becomes $θ = π / 4$ , because $tan (π / 4) = 1$ . Once the limits are values of $θ$ , there is no triangle to draw and no return to $x$ .

When the θ-integral is a secant

A tangent or secant substitution often leaves a power of $sec θ$ to integrate. The simplest case is the integral of $sec θ$ itself. For $\int d x / \sqrt{x^{2} + 4}$ , the substitution $x = 2 tan θ$ leaves exactly that, and the triangle turns the result into $ln (x + \sqrt{x^{2} + 4}) + C$ , with a constant $- ln 2$ absorbed into $C$ .

\int sec θ d θ = ln | sec θ + tan θ | + C

Complete the square first

A quadratic that is not yet a sum or difference of squares becomes one when you complete the square. $x^{2} + 4 x + 13$ $= {(x + 2)}^{2} + 9$ , so the shift $u = x + 2$ gives $\sqrt{u^{2} + 9}$ , and then $u = 3 tan θ$ .

Where it breaks

Try an ordinary substitution first. In $\int x \sqrt{9 - x^{2}} d x$ the factor $x$ is, up to a constant, the derivative of $9 - x^{2}$ , so $u = 9 - x^{2}$ gives $- (1 / 3) {(9 - x^{2})}^{3 / 2} + C$ in one line. Trigonometric substitution is for the integrals where that factor is missing.

Each form has its own substitution, and the wrong one leaves the root in place. $x = 3 sin θ$ in $\sqrt{9 + x^{2}}$ gives $\sqrt{9 + 9 {sin}^{2} θ}$ , which no identity simplifies. The root must hold a quadratic, too. $\sqrt{9 - x^{3}}$ is not a sum or difference of squares, and no trigonometric substitution removes it.

Outside the ranges the absolute value comes back. $\sqrt{{cos}^{2} θ}$ is $∣ cos θ ∣$ , and it equals $cos θ$ only for $θ$ between $- π / 2$ and $π / 2$ . For $\sqrt{x^{2} - a^{2}}$ with $x \leq - a$ , $θ$ lies between $π / 2$ and $π$ , where $tan θ$ is negative, and the root equals $- a tan θ$ . The examples here take $x \geq a$ .

An answer with an inverse trigonometric function carries a domain. $arcsin (x / 3)$ is defined only for $x$ from $- 3$ to $3$ , the same interval on which $\sqrt{9 - x^{2}}$ is real.

Four integrals that look alike
Integral	What it needs	Result
$\int x \sqrt{9 - x^{2}} d x$	$u$ $= 9 - x^{2}$	$- (1 / 3) {(9 - x^{2})}^{3 / 2} + C$
$\int \sqrt{9 - x^{2}} d x$	$x$ $= 3 sin θ$	$(9 / 2) arcsin (x / 3) + (x / 2) \sqrt{9 - x^{2}} + C$
$\int d x / (9 + x^{2})$	$x$ $= 3 tan θ$	$(1 / 3) arctan (x / 3) + C$
$\int d x / \sqrt{x^{2} - 9}$ , $x > 3$	$x$ $= 3 sec θ$	$ln (x + \sqrt{x^{2} - 9}) + C$

Check your answer

An answer here travels into $θ$ and back, so test it where it lands, in $x$ .

Take the derivative back in $x$ . It should return the integrand in $x$ , not the integrand in $θ$ . The derivative of $\sqrt{x^{2} - 9} / (9 x)$ simplifies to $1 / (x^{2} \sqrt{x^{2} - 9})$ , the integrand of the secant example.

Compare a definite value with a number you trust. The first worked example’s answer gives $9 π / 4$ from $0$ to $3$ , the quarter circle. For the tangent example, Slopes, sums & signed area integrates the original integrand in $x$ and gives about $0.1768$ , which is $\sqrt{2} / 8$ .

Check the triangle. Its sides must satisfy the Pythagorean theorem, and the ratio it shows must be the substitution, $sin θ = x / a$ , $tan θ = x / a$ or $sec θ = x / a$ .

In a Math box, Chalk checks the substitution and its differential. Write the integral, then $x = 3 sin θ$ , then $d x = 3 cos θ d θ$ . The differential is checked against the derivative of the substitution. A slip such as $d x = 3 sin θ d θ$ reads ✗ “Doesn’t match dx = x′(θ) dθ.” A θ-integral with its antiderivative on one row is checked too, such as $\int \frac{1}{4} {csc}^{2} θ d θ$ $= - \frac{1}{4} cot θ + C$ . One with ${cos}^{2} θ$ is checked as well, written directly or in its half-angle form.

Chalk also checks the step from the integral in $x$ to the integral in $θ$ . A θ-integral with the $d x$ factor left out, $\int 3 cos θ d θ$ $= 3 sin θ + C$ under $\int \sqrt{9 - x^{2}} d x$ , reads ✗, because with $x$ written out it is not the integral in row 1. Count the factors in the θ-integral: what the root becomes, times what $d x$ becomes.

Write the first row as $\int 1 / (x^{2} \sqrt{x^{2} - 9}) d x$ , with $d x$ after the fraction. When every step is right, the answer in $x$ at the bottom, $\sqrt{x^{2} - 9} / (9 x) + C$ , reads ✓ against it. A wrong answer there, such as $1 / (3 x) + C$ , reads ✗. Written with $d x$ on top of the fraction, the integral is not read yet.

Some rows of this method read “?”, which means not verified, not wrong. An answer with $arccos$ reads ✓ “Equal for x > 2.” when it stands straight under its integral, and “?” when it comes after the substitution rows. The row $\int sec θ d θ$ by itself below the substitution rows reads “?”, but $\int sec θ d θ$ $= ln ∣ sec θ + tan θ ∣ + C$ on one row reads ✓. A root in the denominator is read too: the method table’s own answer, $- \sqrt{4 - x^{2}} / (4 x) + C$ , reads ✓. The row that rewrites an integral in $x$ as one in $θ$ with new limits reads ✓ when the limits carry over. Keeping the old limits reads ✗. The θ-integral and its value can share one row, as the tangent example’s box has them.

A sector, a triangle and the hyperbolic route

The answer to $\int \sqrt{9 - x^{2}} d x$ has a picture. From $0$ to $x$ , the region under the arc of radius $3$ splits along the radius into a sector and a triangle. The sector, with angle $θ = arcsin (x / 3)$ at the center, has area $(9 / 2) θ$ . The triangle, with base $x$ and height $\sqrt{9 - x^{2}}$ , has area $(x / 2) \sqrt{9 - x^{2}}$ . At $x = 1.5$ they are about $2.356$ and $1.949$ , which add to $4.305$ , the area under the arc from $0$ to $1.5$ .

Hyperbolic functions give a second family of substitutions. The identity ${cosh}^{2} t - {sinh}^{2} t = 1$ plays the part of the Pythagorean identity, so $x = a sinh t$ removes $\sqrt{a^{2} + x^{2}}$ with no secant at all. It turns $\int d x / \sqrt{x^{2} + a^{2}}$ into $\int d t = t$ , and $t = arcsinh (x / a)$ $= ln (x + \sqrt{x^{2} + a^{2}}) - ln a$ , the answer the secant route gives.

Some integrals lead to $\int {sec}^{3} θ d θ$ , as the arc length of a parabola does. Integration by parts with $u = sec θ$ and $d v = {sec}^{2} θ d θ$ brings the integral back, and solving for it gives $\frac{1}{2} (sec θ tan θ + ln ∣ sec θ + tan θ ∣)$ $+ C$ .

\int_{0}^{x} \sqrt{9 - t^{2}} d t = \frac{9}{2} arcsin \frac{x}{3} + \frac{x}{2} \sqrt{9 - x^{2}}

\begin{aligned} \int_{0}^{x} \sqrt{9 - t^{2}} d t \\ = \frac{9}{2} arcsin \frac{x}{3} + \frac{x}{2} \sqrt{9 - x^{2}} \end{aligned}

Common mistakes

Replacing the root and writing $d θ$ for $d x$ . With $x = 3 sin θ$ , $d x$ is $3 cos θ d θ$ , and leaving out the $3 cos θ$ turns the quarter circle’s $9 π / 4$ into $3$ .
Dropping a when it is not $1$ . $\sqrt{9 - x^{2}}$ needs $x = 3 sin θ$ , because $x = sin θ$ leaves $\sqrt{9 - {sin}^{2} θ}$ , which no identity simplifies.
Leaving the answer in $θ$ . An indefinite integral in $x$ ends in $x$ , so read the trigonometric functions of $θ$ from the triangle and write $θ$ itself as $arcsin (x / a)$ , $arctan (x / a)$ or $arccos (a / x)$ .
Reading the wrong ratio from the triangle. With $x = 3 sec θ$ the hypotenuse is $x$ and the adjacent side is $3$ , so $cos θ = 3 / x$ and $sin θ = \sqrt{x^{2} - 9} / x$ , not $3 / x$ .
Putting the x-limits on the θ-integral. With $x = 2 tan θ$ , the limits $x = 0$ and $x = 2$ become $θ = 0$ and $θ = π / 4$ , and keeping $0$ and $2$ gives a wrong number.
Writing $\sqrt{{cos}^{2} θ}$ as $cos θ$ without a reason. It is $∣ cos θ ∣$ , and it equals $cos θ$ only because $θ$ is kept between $- π / 2$ and $π / 2$ , where the cosine is not negative.
Using a trigonometric substitution where $u = a^{2} - x^{2}$ would do. In $\int x \sqrt{9 - x^{2}} d x$ the factor $x$ is half the derivative of $9 - x^{2}$ , so the ordinary substitution finishes in one line.

Key terms

Trigonometric substitution: A method for integrals that contain $\sqrt{a^{2} - x^{2}}$ , $\sqrt{a^{2} + x^{2}}$ or $\sqrt{x^{2} - a^{2}}$ . Set $x = a sin θ$ , $x = a tan θ$ or $x = a sec θ$ , so that a Pythagorean identity turns the expression under the root into a perfect square. Replace $d x$ too, and use a right triangle to convert the answer back to $x$ .
Integration by substitution: A method that reverses the chain rule: when an integrand has the form $f (g (x)) \cdot g^{'} (x)$ , set $u = g (x)$ , so $d u = g^{'} (x) d x$ , and integrate $f (u)$ instead. A definite integral also converts its limits to u-values.
Pythagorean identity: The identity ${sin}^{2} θ + {cos}^{2} θ = 1$ , true for every angle because $(cos θ, sin θ)$ lies on the unit circle. Dividing it by ${cos}^{2} θ$ or by ${sin}^{2} θ$ gives $1 + {tan}^{2} θ = {sec}^{2} θ$ and $1 + {cot}^{2} θ = {csc}^{2} θ$ .
Inverse sine: $arcsin x$ , or ${sin}^{- 1} x$ : the angle between $- π / 2$ and $π / 2$ ( $- 90 °$ to $90 °$ ) whose sine is $x$ , for $- 1 \leq x \leq 1$ . The equation $sin θ = x$ has other solutions too, and $arcsin$ gives just this one.
Inverse tangent: $arctan x$ , or ${tan}^{- 1} x$ : the angle between $- π / 2$ and $π / 2$ ( $- 90 °$ to $90 °$ ) whose tangent is $x$ . It is defined for every real $x$ .
Antiderivative: A function whose derivative is the given function, such as $x^{3}$ for $3 x^{2}$ . Any two antiderivatives on one interval differ by a constant, which is why answers carry $+ C$ .