Trigonometric substitution — x = a sin θ, a tan θ, a sec θ cases: Question bank

Definition Setup and notation — read first

Throughout this page, $a$ is a fixed positive constant ( $a > 0$ ) that comes from the integrand (e.g. $9 - x^{2}$ has $a = 3$ ), while $x$ is the variable of integration and $θ$ is the new angle variable we introduce. The three radicals and their substitutions, with the full $θ$ -range for each, are:

Radical	Substitution	Root becomes	$θ$ -range (why)
$a^{2} - x^{2}$	$x = a sin θ$	$a cos θ$	$[- \frac{π}{2}, \frac{π}{2}]$ so $cos θ \geq 0$
$a^{2} + x^{2}$	$x = a tan θ$	$a sec θ$	$(- \frac{π}{2}, \frac{π}{2})$ so $sec θ > 0$
$x^{2} - a^{2}$	$x = a sec θ$	$a tan θ$	$[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})$ so $tan θ \geq 0$

Each range is chosen for exactly one reason: it keeps the trig factor the root simplifies to non-negative, so $u^{2} = ∣ u ∣ = u$ with no sign trap. See the reference triangles in Figure s01.

Figure — Trigonometric substitution — x = a sin θ, a tan θ, a sec θ cases

The three reference triangles above are the whole "back-substitution" engine: whichever side you called $x$ , the picture hands you every trig function of $θ$ in terms of $x$ and $a$ . Keep them in view while answering the cards.

True or false — justify

You can substitute

x = a cos θ

instead of

x = a sin θ

for

a^{2} - x^{2}

True —

1 - cos^{2} θ = sin^{2} θ

also kills the root, giving

a sin θ

. Sine is just the conventional choice because it keeps the clean

θ

-range

[- \frac{π}{2}, \frac{π}{2}]

where the correspondence

x \leftrightarrow θ

is one-to-one.

a^{2} cos^{2} θ = a cos θ

is always valid.

False in general —

u^{2} = ∣ u ∣

, so it equals

a ∣ cos θ ∣

(recall

a > 0

). It becomes

a cos θ

only because the chosen range

[- \frac{π}{2}, \frac{π}{2}]

forces

cos θ \geq 0

Trig substitution requires the radical to be present in the integrand.

False — a denominator like

(x^{2} + 7)^{2}

with no visible root still calls for

x = 7 tan θ

. The reason is structural:

x^{2} + 7

is a sum

a^{2} + x^{2}

with

a = 7

, an irreducible quadratic;

x = 7 tan θ

turns

x^{2} + 7

into

7 sec^{2} θ

, so the whole

()^{2}

becomes

49 sec^{4} θ

and the integral collapses to a clean power of

cos θ

. The surd sign was never the deciding feature — the

+

-structure is.

After substituting, an answer left in terms of

θ

is a complete solution.

False — the integral was in

x

, so the answer must be too. You must back-substitute via the reference triangle (Figure s01); leaving

θ

is an unfinished problem.

For

x^{2} + a^{2}

the substitution

x = a sec θ

also works.

False —

a^{2} sec^{2} θ + a^{2} = a^{2} (sec^{2} θ + 1)

, which is not a perfect square (there is no identity for

sec^{2} + 1

), so the root doesn't simplify. The plus-structure demands

tan

, because

1 + tan^{2} θ = sec^{2} θ

The

d x

factor is optional bookkeeping you can restore at the end.

False —

d x = a cos θ d θ

(etc.) is part of the integrand and usually cancels crucial factors; skipping it changes the integral entirely.

\int \frac{d x}{a ^{2} + x ^{2}} = \frac{1}{a} arctan \frac{x}{a} + C

cannot be derived without a full trig substitution.

False. It can be gotten straight from the known derivative

\frac{d}{d x} arctan \frac{x}{a} = \frac{a}{a ^{2} + x ^{2}}

, no substitution needed. Trig sub (

x = a tan θ

, collapsing it to

\frac{1}{a} \int d θ

) is one clean route, and it explains where the

\frac{1}{a}

comes from — the

d x = a sec^{2} θ d θ

over

a^{2} sec^{2} θ

cancellation — but it is not required.

Spot the error

9 - x^{2} = 3 cos θ

, and since it's a length it's positive, so no range check is needed."

The positivity is not automatic from "it's a length" — it follows from the range

θ \in [- \frac{π}{2}, \frac{π}{2}]

making

cos θ \geq 0

. The range restriction is the actual justification, and you must state it.

"For

x^{2} - 25

I set

x = 5 sin θ

, giving

25 sin^{2} θ - 25 = 5 sin^{2} θ - 1

The error is the wrong sub —

sin^{2} θ - 1 \leq 0

, so you get an imaginary root. The

x^{2} - a^{2}

structure needs

x = 5 sec θ

, since

sec^{2} θ - 1 = tan^{2} θ \geq 0

\int 4 - x^{2} d x = 4 \int cos^{2} θ d θ

after

x = 2 sin θ

Here

a = 2

: the root gives

2 cos θ

and

d x = 2 cos θ d θ

, so the product is

2 cos θ \cdot 2 cos θ = 4 cos^{2} θ

— actually correct. Trap: verify the coefficient is genuinely

a^{2} = 2^{2} = 4

rather than fudging a number to make it "look right."

"After

x = tan θ

, I got

tan θ - θ + C

, and since

tan θ = x^{2} - 1

I'm done."

Mixed cases —

tan θ = x

came from a

tan

sub, but

tan θ = x^{2} - 1

belongs to a

sec

sub. You must build the triangle from your own substitution, not borrow another's (compare the two triangles in Figure s01).

θ = arcsin \frac{x}{a}

is valid for all real

x

Only for

∣ x ∣ \leq a

arcsin

is defined on

[- 1, 1]

, which matches

a^{2} - x^{2}

being real exactly when

∣ x ∣ \leq a

— the domain and the substitution agree, which is the point.

"For

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

I can pull

x^{2}

out as a constant."

x^{2}

is a variable, not a constant — it becomes

4 tan^{2} θ

after substitution and stays inside the integral. Only genuine constants (

a^{2} = 4

) may leave.

Why questions

Why do the Pythagorean identities make trig substitution work at all?

They convert a sum or difference of two squares into a single squared term, so

single square

becomes a bare trig function with no root left.

Why is

tan

(not

sin

sec

) the right sub for

a^{2} + x^{2}

Only

1 + tan^{2} θ = sec^{2} θ

turns a plus of squares into one square;

sin

cos

handle differences, and

sec

handles

x^{2} - a^{2}

Why does the

sec

case have a split range

[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})

instead of one interval?

x = a sec θ

must reach both

x \geq a

and

x \leq - a

, and we need

tan θ \geq 0

on each piece so the root

a tan θ

stays positive; one continuous interval can't do both.

Why do we build a reference right triangle at the end?

The substitution defined

θ

via a ratio of sides (e.g.

sin θ = x / a

), so the triangle (Figure s01) lets us read off every trig function of

θ

in terms of

x

to undo the substitution. See Reference right triangle method.

Why does power-reduction appear so often in these integrals?

Killing the root frequently leaves

cos^{2} θ

sin^{2} θ

, which aren't directly integrable; the double-angle identity linearises them into

\frac{1 \pm c o s 2 θ}{2}

. See Power-reduction & double-angle formulas.

Why prefer trig sub over partial fractions for

\frac{1}{( x ^{2} + a ^{2} ) ^{2}}

Partial fractions can't split an irreducible quadratic power over the reals;

x = a tan θ

turns it into a clean

cos^{2}

-type integral instead. Compare Partial fractions.

Why does the same triangle idea reappear in arc-length problems?

Arc length integrals produce

1 + (f^{'})^{2}

, exactly the

a^{2} + x^{2}

shape, so trig (or Hyperbolic substitution) sub is the natural tool. See Arc length and surface area.

Edge cases

What happens to

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

when

x = a

(the endpoint)?

The root is

0

and

arcsin \frac{a}{a} = \frac{π}{2}

, so the formula gives

\frac{a ^{2}}{2} \cdot \frac{π}{2}

— a finite value; the integrand touches zero smoothly with no blow-up.

What if

a = 0

a^{2} - x^{2}

It degenerates to

- x^{2}

, real only at

x = 0

; the substitution collapses since there's no triangle to build. Trig sub assumes

a > 0

, which is why we fixed

a > 0

at the top.

For

x^{2} - a^{2}

, what goes wrong at

x = a

exactly?

The root is

0

and

θ = 0

(since

sec 0 = 1

), a boundary of the allowed range; the integrand

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

is fine, but

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

would be singular there.

Can trig sub handle

x^{2} + a^{2}

when

x < 0

Yes —

x = a tan θ

with

θ \in (- \frac{π}{2}, 0)

covers negative

x

, and

sec θ > 0

throughout, so the root stays positive with no sign issue.

Is trig sub the only way to do

x^{2} + a^{2}

No — hyperbolic substitution

x = a sinh t

uses

1 + sinh^{2} t = cosh^{2} t

and often avoids messy back-substitution; it's a genuine alternative. See Hyperbolic substitution.

What if the quadratic isn't already

a^{2} \pm x^{2}

, e.g.

x^{2} + 4 x + 13

Complete the square first:

x^{2} + 4 x + 13 = (x + 2)^{2} + 9

, then substitute

x + 2 = 3 tan θ

(so effectively

a = 3

). Trig sub needs the standard form; algebra sets it up.

For

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

, why is the answer

arcsin

but

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

gives a log?

The

sin

-case collapses to

\int d θ = θ = arcsin

, but the

sec

-case leaves

\int sec θ d θ

, whose antiderivative is a logarithm — different residual integrals, different function families.

Connections

Pythagorean identities — why every case simplifies.
Integration by substitution (u-sub) — the change-of-variable backbone.
Reference right triangle method — the back-substitution tool (drawn in Figure s01).
Power-reduction & double-angle formulas — for the leftover $sin^{2}, cos^{2}$ .
Partial fractions — the alternative when the quadratic is reducible.
Arc length and surface area — where these roots naturally arise.
Hyperbolic substitution — the parallel toolkit.