Blog · Calculus guides · 29 October 2026 · 4 min read

Trig Substitution: Steps, Examples & Practice Problems

Q: How do you know which trig substitution to use?

By the sign and the order: a^2 - x^2 takes x = a sinθ, a^2 + x^2 takes x = a tanθ, and x^2 - a^2 takes x = a secθ. Each is picked so a Pythagorean identity turns the expression into a perfect square.

Q: Why do you draw a triangle in trig substitution?

To convert the θ answer back into x. The substitution tells you one trig ratio, the triangle and Pythagoras give you all the others.

Q: Do I need trig substitution for 1/(√(a^2 - x^2))?

Not if you know the result: ∫ dx/(√(a^2 - x^2)) = arcsinx/a + C. Trig substitution is how that formula is proved, so it is a good one to derive once and then memorize.

Q: Can I use hyperbolic substitution instead?

Yes: x = a sinh u handles √(a^2 + x^2) neatly, which is why some answers come out with sinh^(-1) in them. It agrees with the trig version up to a constant.

Trig substitution replaces $x$ with a trig function so that a square root of a sum or difference of squares collapses into a single trig term. Use $x = a sin θ$ for $a^{2} - x^{2}$ , $x = a tan θ$ for $a^{2} + x^{2}$ , and $x = a sec θ$ for $x^{2} - a^{2}$ . Integrate in $θ$ , then draw a right triangle to convert back to $x$ .

The formula

a^{2} - x^{2} a^{2} + x^{2} x^{2} - a^{2} x = a sin θ x = a tan θ x = a sec θ 1 - sin^{2} θ = cos^{2} θ 1 + tan^{2} θ = sec^{2} θ sec^{2} θ - 1 = tan^{2} θ

Three shapes, three substitutions, and each one is a Pythagorean identity wearing a disguise.

When to use it

Reach for trig substitution when the integrand contains $a^{2} - x^{2}$ , $a^{2} + x^{2}$ , $x^{2} - a^{2}$ or $a^{2} + x^{2}$ in a denominator, and there is no stray $x$ outside that would make u-substitution work. That last check matters: $\int x 9 - x^{2} d x$ looks like a trig substitution but is a two-line u-substitution with $u = 9 - x^{2}$ . Try the cheap method first and bring out the triangles only if it fails. Completing the square can also unlock it: $x^{2} + 4 x + 13$ is $(x + 2)^{2} + 9$ , which is the tangent shape with $a = 3$ once you let $u = x + 2$ .

How to do it, step by step

Match the shape. Find $a$ and pick the substitution from the table: minus with $a^{2}$ first is sine, plus is tangent, minus with $x^{2}$ first is secant.
Substitute everything. Replace $x$ , replace $d x$ (for $x = a sin θ$ , $d x = a cos θ d θ$ ), and simplify the root with the identity.
Integrate in $θ$ . What is left is a trig integral, often $cos^{2} θ$ , $sec^{2} θ$ or $tan^{2} θ$ , which the double-angle or Pythagorean identities finish off.
Draw the triangle and convert back. From $x = a sin θ$ , label the opposite side $x$ and the hypotenuse $a$ ; Pythagoras gives the third side. Read every trig function of $θ$ off it, and write $θ$ itself as an inverse trig function of $x$ .

Worked examples

Example 1: A tangent substitution

Easy

Integrate:

\int \frac{1}{x ^{2} + 4} d x

The shape is $a^{2} + x^{2}$ with $a = 2$ , so let $x = 2 tan θ$ . Then $d x = 2 sec^{2} θ d θ$ and $x^{2} + 4 = 4 tan^{2} θ + 4 = 4 sec^{2} θ$ .

\int \frac{2 sec ^{2} θ}{4 sec ^{2} θ} d θ = \int \frac{1}{2} d θ = \frac{θ}{2} + C

Everything cancels, which is the substitution doing its job: the $sec^{2} θ$ from $d x$ wipes out the $sec^{2} θ$ the identity made. Since $tan θ = \frac{x}{2}$ , $θ = arctan \frac{x}{2}$ .

This one is worth remembering as a pattern, because it comes up constantly: $\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} arctan \frac{x}{a} + C$ . Once you know it, you can skip the substitution entirely.

Answer

\frac{1}{2} arctan (\frac{x}{2}) + C

Example 2: A sine substitution and a triangle

Medium

Integrate:

\int 9 - x^{2} d x

The shape is $a^{2} - x^{2}$ with $a = 3$ , so let $x = 3 sin θ$ . Then $d x = 3 cos θ d θ$ and $9 - 9 sin^{2} θ = 3 cos θ$ .

\int 3 cos θ \cdot 3 cos θ d θ = 9 \int cos^{2} θ d θ = \frac{9}{2} (θ + sin θ cos θ) + C

That used $cos^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ and $sin 2 θ = 2 sin θ cos θ$ . Now the triangle: opposite $x$ , hypotenuse $3$ , so the adjacent side is $9 - x^{2}$ .

sin θ = \frac{x}{3}, cos θ = \frac{9 - x ^{2}}{3}, θ = arcsin \frac{x}{3}

Substitute back: $\frac{9}{2} sin θ cos θ = \frac{9}{2} \cdot \frac{x 9 - x ^{2}}{9} = \frac{x}{2} 9 - x^{2}$ .

A sanity check: between $- 3$ and $3$ , this integral is the area of a half-disk of radius $3$ . Plugging in the limits gives $\frac{9}{2} (\frac{π}{2} + \frac{π}{2}) = \frac{9 π}{2}$ , which is exactly half of $π \cdot 3^{2}$ .

Answer

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

Example 3: A secant substitution

Harder

Integrate, for $x > 2$ :

\int \frac{x ^{2} - 4}{x} d x

The shape is $x^{2} - a^{2}$ with $a = 2$ , so let $x = 2 sec θ$ . Then $d x = 2 sec θ tan θ d θ$ and $4 sec^{2} θ - 4 = 2 tan θ$ .

\int \frac{2 tan θ}{2 sec θ} \cdot 2 sec θ tan θ d θ = 2 \int tan^{2} θ d θ = 2 \int (sec^{2} θ - 1) d θ = 2 tan θ - 2 θ + C

Triangle: $sec θ = \frac{x}{2}$ means hypotenuse $x$ and adjacent side $2$ , so the opposite side is $x^{2} - 4$ and $tan θ = \frac{x ^{2} - 4}{2}$ .

So $2 tan θ = x^{2} - 4$ , and $θ$ is the angle whose tangent is $\frac{x ^{2} - 4}{2}$ . You will also see it written as $arcsec \frac{x}{2}$ ; for $x > 2$ they are the same angle.

The $tan^{2} θ$ step is the one to watch. There is no direct antiderivative of $tan^{2} θ$ in most tables, but $sec^{2} θ$ integrates to $tan θ$ , so the identity $tan^{2} θ = sec^{2} θ - 1$ swaps an awkward integral for two easy ones.

Answer

x^{2} - 4 - 2 arctan (\frac{x ^{2} - 4}{2}) + C

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Integrate:

\int \frac{5}{16 - x ^{2}} d x

Problem 2

Medium

Integrate:

\int \frac{x ^{2}}{4 - x ^{2}} d x

Problem 3

Harder

Integrate, for $x > 0$ :

\int \frac{1}{x ^{2} x ^{2} + 4} d x

Common mistakes

Forgetting to substitute $d x$ . Replacing $x$ but leaving $d x$ alone loses a whole factor, usually the one that was going to cancel.
Picking the wrong substitution. $x^{2} - 9$ is secant, not sine. Look at which term comes first under the root: the constant first means sine, $x^{2}$ first means secant.
Leaving $θ$ in the final answer. The question was in $x$ , so the answer must be too. The triangle exists for exactly this step.
Losing the $a$ . With $x = 3 sin θ$ , $θ = arcsin \frac{x}{3}$ , not $arcsin x$ . The $a$ has to go back in when you undo the substitution.
Using trig substitution when u-substitution works. If there is an $x$ outside the root that matches the derivative of what is inside, $u =$ the inside is far quicker.

Practice it now

No clock, three strikes

Trig substitution turns up in Diverge from about d9: roots of $a^{2} - x^{2}$ and $x^{2} + a^{2}$ , and $a^{2} + x^{2}$ in a denominator. There are no secant ( $x^{2} - a^{2}$ ) problems in the game yet, and answers are checked up to a constant, so no $+ C$ is needed.

Play Diverge See it on the formula sheet

FAQ

How do you know which trig substitution to use?: By the sign and the order: $a^{2} - x^{2}$ takes $x = a sin θ$ , $a^{2} + x^{2}$ takes $x = a tan θ$ , and $x^{2} - a^{2}$ takes $x = a sec θ$ . Each is picked so a Pythagorean identity turns the expression into a perfect square.
Is trig substitution the same as u-substitution?: It is a cousin. U-substitution replaces an inner function with $u$ ; trig substitution goes the other way and replaces $x$ itself with a function of $θ$ . Both need $d x$ rewritten.
Why do you draw a triangle in trig substitution?: To convert the $θ$ answer back into $x$ . The substitution tells you one trig ratio, the triangle and Pythagoras give you all the others.
Do I need trig substitution for $\frac{1}{a ^{2} - x ^{2}}$ ?: Not if you know the result: $\int \frac{d x}{a ^{2} - x ^{2}} = arcsin \frac{x}{a} + C$ . Trig substitution is how that formula is proved, so it is a good one to derive once and then memorize.
Can I use hyperbolic substitution instead?: Yes: $x = a sinh u$ handles $a^{2} + x^{2}$ neatly, which is why some answers come out with $sinh^{- 1}$ in them. It agrees with the trig version up to a constant.
Where can I find trig substitution practice problems?: Right on this page: "Your turn" has three trig substitution practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, Diverge, which has no clock, is free and needs no account.

When to use it

Reach for trig substitution when the integrand contains

a^{2} - x^{2}

a^{2} + x^{2}

x^{2} - a^{2}

a^{2} + x^{2}

in a denominator, and there is no stray

x

outside that would make u-substitution work. That last check matters:

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

looks like a trig substitution but is a two-line u-substitution with

u = 9 - x^{2}

. Try the cheap method first and bring out the triangles only if it fails. Completing the square can also unlock it:

x^{2} + 4 x + 13

(x + 2)^{2} + 9

, which is the tangent shape with

a = 3

once you let

u = x + 2

How to do it, step by step

Match the shape. Find

a

and pick the substitution from the table: minus with

a^{2}

first is sine, plus is tangent, minus with

x^{2}

first is secant.

Substitute everything. Replace

x

, replace

d x

(for

x = a sin θ

d x = a cos θ d θ

), and simplify the root with the identity.

Integrate in $θ$ . What is left is a trig integral, often

cos^{2} θ

sec^{2} θ

tan^{2} θ

, which the double-angle or Pythagorean identities finish off.

Draw the triangle and convert back. From

x = a sin θ

, label the opposite side

x

and the hypotenuse

a

; Pythagoras gives the third side. Read every trig function of

θ

off it, and write

θ

itself as an inverse trig function of

x

Worked examples

Example 1: A tangent substitution

Easy

Integrate:

\int \frac{1}{x ^{2} + 4} d x

The shape is $a^{2} + x^{2}$ with $a = 2$ , so let $x = 2 tan θ$ . Then $d x = 2 sec^{2} θ d θ$ and $x^{2} + 4 = 4 tan^{2} θ + 4 = 4 sec^{2} θ$ .

\int \frac{2 sec ^{2} θ}{4 sec ^{2} θ} d θ = \int \frac{1}{2} d θ = \frac{θ}{2} + C

Everything cancels, which is the substitution doing its job: the $sec^{2} θ$ from $d x$ wipes out the $sec^{2} θ$ the identity made. Since $tan θ = \frac{x}{2}$ , $θ = arctan \frac{x}{2}$ .

Answer

\frac{1}{2} arctan (\frac{x}{2}) + C

Example 2: A sine substitution and a triangle

Medium

Integrate:

\int 9 - x^{2} d x

The shape is $a^{2} - x^{2}$ with $a = 3$ , so let $x = 3 sin θ$ . Then $d x = 3 cos θ d θ$ and $9 - 9 sin^{2} θ = 3 cos θ$ .

\int 3 cos θ \cdot 3 cos θ d θ = 9 \int cos^{2} θ d θ = \frac{9}{2} (θ + sin θ cos θ) + C

That used $cos^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ and $sin 2 θ = 2 sin θ cos θ$ . Now the triangle: opposite $x$ , hypotenuse $3$ , so the adjacent side is $9 - x^{2}$ .

sin θ = \frac{x}{3}, cos θ = \frac{9 - x ^{2}}{3}, θ = arcsin \frac{x}{3}

Substitute back: $\frac{9}{2} sin θ cos θ = \frac{9}{2} \cdot \frac{x 9 - x ^{2}}{9} = \frac{x}{2} 9 - x^{2}$ .

Answer

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

Example 3: A secant substitution

Harder

Integrate, for $x > 2$ :

\int \frac{x ^{2} - 4}{x} d x

The shape is $x^{2} - a^{2}$ with $a = 2$ , so let $x = 2 sec θ$ . Then $d x = 2 sec θ tan θ d θ$ and $4 sec^{2} θ - 4 = 2 tan θ$ .

\int \frac{2 tan θ}{2 sec θ} \cdot 2 sec θ tan θ d θ = 2 \int tan^{2} θ d θ = 2 \int (sec^{2} θ - 1) d θ = 2 tan θ - 2 θ + C

Triangle: $sec θ = \frac{x}{2}$ means hypotenuse $x$ and adjacent side $2$ , so the opposite side is $x^{2} - 4$ and $tan θ = \frac{x ^{2} - 4}{2}$ .

So $2 tan θ = x^{2} - 4$ , and $θ$ is the angle whose tangent is $\frac{x ^{2} - 4}{2}$ . You will also see it written as $arcsec \frac{x}{2}$ ; for $x > 2$ they are the same angle.

Answer

x^{2} - 4 - 2 arctan (\frac{x ^{2} - 4}{2}) + C

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Integrate:

\int \frac{5}{16 - x ^{2}} d x

Problem 2

Medium

Integrate:

\int \frac{x ^{2}}{4 - x ^{2}} d x

Problem 3

Harder

Integrate, for $x > 0$ :

\int \frac{1}{x ^{2} x ^{2} + 4} d x

Common mistakes

Forgetting to substitute $d x$ . Replacing

x

but leaving

d x

alone loses a whole factor, usually the one that was going to cancel.

Picking the wrong substitution.

x^{2} - 9

is secant, not sine. Look at which term comes first under the root: the constant first means sine,

x^{2}

first means secant.

Leaving $θ$ in the final answer. The question was in

x

, so the answer must be too. The triangle exists for exactly this step.

Losing the $a$ . With

x = 3 sin θ

θ = arcsin \frac{x}{3}

, not

arcsin x

. The

a

has to go back in when you undo the substitution.

Using trig substitution when u-substitution works. If there is an

x

outside the root that matches the derivative of what is inside,

u =

the inside is far quicker.

Practice it now

No clock, three strikes

Play Diverge See it on the formula sheet

FAQ

How do you know which trig substitution to use?

By the sign and the order:

a^{2} - x^{2}

takes

x = a sin θ

a^{2} + x^{2}

takes

x = a tan θ

, and

x^{2} - a^{2}

takes

x = a sec θ

. Each is picked so a Pythagorean identity turns the expression into a perfect square.

Is trig substitution the same as u-substitution?

It is a cousin. U-substitution replaces an inner function with

u

; trig substitution goes the other way and replaces

x

itself with a function of

θ

. Both need

d x

rewritten.

Why do you draw a triangle in trig substitution?

To convert the

θ

answer back into

x

. The substitution tells you one trig ratio, the triangle and Pythagoras give you all the others.

Do I need trig substitution for

\frac{1}{a ^{2} - x ^{2}}

Not if you know the result:

\int \frac{d x}{a ^{2} - x ^{2}} = arcsin \frac{x}{a} + C

. Trig substitution is how that formula is proved, so it is a good one to derive once and then memorize.

Can I use hyperbolic substitution instead?

Yes:

x = a sinh u

handles

a^{2} + x^{2}

neatly, which is why some answers come out with

sinh^{- 1}

in them. It agrees with the trig version up to a constant.

Where can I find trig substitution practice problems?

Right on this page: "Your turn" has three trig substitution practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, Diverge, which has no clock, is free and needs no account.

Trig Substitution: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A tangent substitution

Example 2: A sine substitution and a triangle

Example 3: A secant substitution

Your turn

Common mistakes

FAQ

Keep going

Trig Substitution: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A tangent substitution

Example 2: A sine substitution and a triangle

Example 3: A secant substitution

Your turn

Common mistakes

FAQ

Keep going