Integrals That Give Inverse Trig Functions

Integrals that give inverse trig functions have $\sqrt{a^2 - u^2}$ or $a^2 + u^2$ in the denominator and a constant on top. The two main rules are $\int \dfrac{1}{\sqrt{a^2 - u^2}}\,du = \arcsin\dfrac{u}{a} + C$ and $\int \dfrac{1}{a^2 + u^2}\,du = \dfrac{1}{a}\arctan\dfrac{u}{a} + C$ . A third rule gives arcsecant.

Before this lesson: Derivatives of Inverse Trig Functions, U-Substitution and Integrals That Give Natural Logarithms

Where these integrals come from

Each rule is a derivative read backward. You learned that (arcsin x)′ = 1/√(1 − x²) and (arctan x)′ = 1/(1 + x²). So the integrals of those two fractions are arcsin x + C and arctan x + C.

The number a stretches the pattern. Factor a² out of the denominator and substitute u/a, and you get the a-versions of the rules below. The area picture shows the simplest case.

Graph of y = 1/(1 + x squared) for x from −3 to 3, a bell-shaped curve with peak 1 at x = 0. The region under the curve from x = 0 to x = 1 is shaded and labeled area = arctan 1 = π/4. — The area under y = 1/(1 + x²) from 0 to 1 is arctan 1 − arctan 0 = π/4, about 0.785.

Inverse trig integral formulas

For a positive constant a:

\int \dfrac{du}{\sqrt{a^2 - u^2}} = \arcsin\dfrac{u}{a} + C

\int \dfrac{du}{a^2 + u^2} = \dfrac{1}{a}\arctan\dfrac{u}{a} + C

\int \dfrac{du}{u\sqrt{u^2 - a^2}} = \dfrac{1}{a}\operatorname{arcsec}\dfrac{|u|}{a} + C

Arccosine, arccotangent and arccosecant don’t need their own rules. Their derivatives are just the negatives of the ones above.

How to use the rules:

Check the top. It must be a constant (or the du from a substitution), not the derivative of the bottom. If it is the derivative of the bottom, use ln instead.
Match the bottom to a² − u² under a root, a² + u², or u√(u² − a²). Find a and u.
If u is not just x, find du and fix any constant.
Apply the rule. Remember the 1/a in the arctan and arcsec rules.

Inverse trig integral examples

Example 1: Arcsine

Find $\int \dfrac{1}{\sqrt{9 - x^2}}\,dx$ .

9 - x^2 = 3^2 - x^2

Match the form a² − u² with a = 3 and u = x.

\int \dfrac{1}{\sqrt{9 - x^2}}\,dx = \arcsin\dfrac{x}{3} + C

Apply the arcsine rule.

Answer: $\arcsin\dfrac{x}{3} + C$ . Check: the derivative is $\dfrac{1/3}{\sqrt{1 - x^2/9}}$ , which simplifies to $\dfrac{1}{\sqrt{9 - x^2}}$ .

Example 2: Arctangent with a substitution

Find $\int \dfrac{1}{1 + 9x^2}\,dx$ .

1 + 9x^2 = 1^2 + (3x)^2

Match a² + u² with a = 1 and u = 3x.

u = 3x

Let u = 3x.

du = 3\,dx

Differentiate u, so dx = du/3 (see u-substitution).

\int \dfrac{1}{1 + 9x^2}\,dx = \int \dfrac{1}{3}\cdot\dfrac{1}{1 + u^2}\,du

Rewrite in u.

= \dfrac{1}{3}\arctan u + C

Apply the arctangent rule with a = 1.

= \dfrac{1}{3}\arctan(3x) + C

Substitute back.

Answer: $\dfrac{1}{3}\arctan(3x) + C$

Example 3: A definite arcsine integral

Evaluate $\int_0^1 \dfrac{1}{\sqrt{4 - x^2}}\,dx$ .

\int_0^1 \dfrac{1}{\sqrt{4 - x^2}}\,dx = \Big[\arcsin\dfrac{x}{2}\Big]_0^1

Arcsine rule with a = 2.

= \arcsin\dfrac{1}{2} - \arcsin 0

Plug in the limits and subtract.

= \dfrac{\pi}{6} - 0 = \dfrac{\pi}{6}

sin(π/6) = 1/2 and sin 0 = 0 (see inverse trig functions).

Answer: $\dfrac{\pi}{6} \approx 0.524$

Example 4: Completing the square

Find $\int \dfrac{1}{x^2 + 4x + 13}\,dx$ .

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

Split 13 into 4 + 9 to complete the square (see completing the square).

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

Factor the perfect square trinomial.

u = x + 2

Let u = x + 2; then du = dx.

\int \dfrac{1}{x^2 + 4x + 13}\,dx = \int \dfrac{1}{u^2 + 9}\,du

Rewrite in u. Here a = 3.

= \dfrac{1}{3}\arctan\dfrac{u}{3} + C

Apply the arctangent rule.

= \dfrac{1}{3}\arctan\dfrac{x + 2}{3} + C

Substitute back.

Answer: $\dfrac{1}{3}\arctan\dfrac{x + 2}{3} + C$

Example 5: A camera following a car

A security camera is 20 meters from a straight road. It turns to follow a car, and its angle changes at a rate of $\theta'(t) = \dfrac{20}{400 + t^2}$ radians per second, where t is the car’s distance past the closest point in meters. How far does the camera turn while the car goes from t = 0 to t = 20?

\int_0^{20} \dfrac{20}{400 + t^2}\,dt = 20\int_0^{20} \dfrac{1}{20^2 + t^2}\,dt

Factor out 20; the bottom is a² + u² with a = 20.

= 20\cdot\dfrac{1}{20}\Big[\arctan\dfrac{t}{20}\Big]_0^{20}

Apply the arctangent rule.

= \arctan 1 - \arctan 0

20 · (1/20) = 1; plug in the limits.

= \dfrac{\pi}{4} - 0 = \dfrac{\pi}{4}

tan(π/4) = 1.

Answer: the camera turns π/4 radians, which is 45°. That makes sense: when the car is 20 m past the closest point and the camera is 20 m from the road, the angle is 45°.

Common mistakes

Mistake	Example of the error	Fix
Forgetting the 1/a	$\int \dfrac{1}{16 + x^2}\,dx = \arctan\dfrac{x}{4} + C$	Arctan needs 1/a in front: $\dfrac{1}{4}\arctan\dfrac{x}{4} + C$ .
Using a² instead of a	$\int \dfrac{1}{9 + x^2}\,dx = \dfrac{1}{9}\arctan\dfrac{x}{9} + C$	a is the square root: a = 3, so $\dfrac{1}{3}\arctan\dfrac{x}{3} + C$ .
Using ln for 1/(1 + x²)	$\int \dfrac{1}{1 + x^2}\,dx = \ln(1 + x^2) + C$	The top is not 2x, so this is $\arctan x + C$ .
Using arcsine for a difference in the wrong order	$\int \dfrac{1}{\sqrt{x^2 - 4}}\,dx = \arcsin\dfrac{x}{2} + C$	Arcsine needs a² − u², with the constant first. This integral needs another method (trig substitution, Calculus 2).
Skipping the du constant	$\int \dfrac{1}{1 + 9x^2}\,dx = \arctan(3x) + C$	du = 3 dx, so the answer is $\dfrac{1}{3}\arctan(3x) + C$ .

Related methods

Split the numerator

If the top has an x term and a constant, split the fraction. For example, $\int \dfrac{x + 1}{x^2 + 1}\,dx = \int \dfrac{x}{x^2 + 1}\,dx + \int \dfrac{1}{x^2 + 1}\,dx$ , which is $\dfrac{1}{2}\ln(x^2 + 1) + \arctan x + C$ . The first piece uses the log rule from integrals that give natural logarithms.

Trig substitution

Roots like $\sqrt{x^2 - 4}$ or $\sqrt{x^2 + 9}$ alone, or powers of a² + x², need trigonometric substitution, a Calculus 2 method.

Practice problems

Find $\int \dfrac{1}{\sqrt{25 - x^2}}\,dx$ .
Find $\int \dfrac{1}{49 + x^2}\,dx$ .
Find $\int \dfrac{1}{\sqrt{1 - 4x^2}}\,dx$ .
Find $\int \dfrac{3}{1 + x^2}\,dx$ .
Evaluate $\int_0^{\sqrt{3}} \dfrac{1}{1 + x^2}\,dx$ .
Evaluate $\int_0^{3/2} \dfrac{1}{\sqrt{9 - x^2}}\,dx$ .
Find $\int \dfrac{e^x}{1 + e^{2x}}\,dx$ .
Find $\int \dfrac{1}{x\sqrt{x^2 - 16}}\,dx$ for x greater than 4.
Find $\int \dfrac{1}{x^2 - 6x + 10}\,dx$ .
A robot arm turns at a rate of $\dfrac{2}{\sqrt{4 - t^2}}$ radians per second. How far does it turn from t = 0 to t = 1 second?

Show answers

\int \dfrac{1}{\sqrt{25 - x^2}}\,dx = \arcsin\dfrac{x}{5} + C

Arcsine rule with a = 5.

Answer: $\arcsin\dfrac{x}{5} + C$

\int \dfrac{1}{49 + x^2}\,dx = \dfrac{1}{7}\arctan\dfrac{x}{7} + C

Arctangent rule with a = 7.

Answer: $\dfrac{1}{7}\arctan\dfrac{x}{7} + C$

u = 2x

1 − 4x² = 1 − (2x)², so a = 1 and u = 2x.

du = 2\,dx

Differentiate u, so dx = du/2.

\int \dfrac{1}{\sqrt{1 - 4x^2}}\,dx = \int \dfrac{1}{2}\cdot\dfrac{1}{\sqrt{1 - u^2}}\,du = \dfrac{1}{2}\arcsin u + C

Rewrite and integrate.

= \dfrac{1}{2}\arcsin(2x) + C

Substitute back.

Answer: $\dfrac{1}{2}\arcsin(2x) + C$

\int \dfrac{3}{1 + x^2}\,dx = 3\arctan x + C

Factor out 3; arctangent rule with a = 1.

Answer: $3\arctan x + C$

\int_0^{\sqrt{3}} \dfrac{1}{1 + x^2}\,dx = \Big[\arctan x\Big]_0^{\sqrt{3}}

Arctangent rule.

= \arctan\sqrt{3} - \arctan 0 = \dfrac{\pi}{3}

tan(π/3) = √3 and tan 0 = 0.

Answer: $\dfrac{\pi}{3}$

\int_0^{3/2} \dfrac{1}{\sqrt{9 - x^2}}\,dx = \Big[\arcsin\dfrac{x}{3}\Big]_0^{3/2}

Arcsine rule with a = 3.

= \arcsin\dfrac{1}{2} - \arcsin 0 = \dfrac{\pi}{6}

(3/2)/3 = 1/2.

Answer: $\dfrac{\pi}{6}$

u = e^x

e²ˣ = (eˣ)², so let u = eˣ.

du = e^x\,dx

The numerator is du.

\int \dfrac{e^x}{1 + e^{2x}}\,dx = \int \dfrac{1}{1 + u^2}\,du = \arctan u + C

Rewrite and integrate.

= \arctan(e^x) + C

Substitute back.

Answer: $\arctan(e^x) + C$

\int \dfrac{1}{x\sqrt{x^2 - 16}}\,dx = \dfrac{1}{4}\operatorname{arcsec}\dfrac{x}{4} + C

Arcsecant rule with a = 4; x is positive, so |x| = x.

Answer: $\dfrac{1}{4}\operatorname{arcsec}\dfrac{x}{4} + C$

x^2 - 6x + 10 = (x - 3)^2 + 1

Complete the square: 10 = 9 + 1.

\int \dfrac{1}{(x - 3)^2 + 1}\,dx = \arctan(x - 3) + C

Let u = x − 3, du = dx; arctangent rule with a = 1.

Answer: $\arctan(x - 3) + C$

10.

\int_0^1 \dfrac{2}{\sqrt{4 - t^2}}\,dt = \Big[2\arcsin\dfrac{t}{2}\Big]_0^1

Factor out 2; arcsine rule with a = 2.

= 2\arcsin\dfrac{1}{2} - 2\arcsin 0 = \dfrac{\pi}{3}

2 · (π/6) = π/3.

Answer: $\dfrac{\pi}{3} \approx 1.047$ radians, or 60°

Inverse trig integrals FAQ

What is the integral of 1/(1 + x²)?

It is arctan x + C. More generally, the integral of 1/(a² + x²) is (1/a) arctan(x/a) + C.

What is the integral of 1/√(1 − x²)?

It is arcsin x + C. With a constant a, the integral of 1/√(a² − x²) is arcsin(x/a) + C.

Why is there no arccos integral rule?

The derivative of arccos x is the negative of the derivative of arcsin x. So −arccos x + C gives the same family of answers as arcsin x + C.

How do you know whether to use ln or arctan?

If the top is the derivative of the bottom, use ln. If the top is a constant and the bottom is a² + u², use arctan.

Summary

$\int \dfrac{du}{\sqrt{a^2 - u^2}} = \arcsin\dfrac{u}{a} + C$
$\int \dfrac{du}{a^2 + u^2} = \dfrac{1}{a}\arctan\dfrac{u}{a} + C$
$\int \dfrac{du}{u\sqrt{u^2 - a^2}} = \dfrac{1}{a}\operatorname{arcsec}\dfrac{|u|}{a} + C$
a is the square root of the constant, not the constant itself.
Complete the square when the bottom is a quadratic with an x term.

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