List of Integrals of Inverse Trigonometric Functions

What Is The List Of Integrals Of Inverse Trigonometric Functions?

The list of integrals of inverse trigonometric functions is a reference table covering two families. The first family integrates the inverse trigonometric functions themselves, such as $\int \arcsin x,dx$; the second family integrates algebraic expressions whose antiderivatives are inverse trig functions, such as $\int \frac{dx}{1+x^2} = \arctan x + C$.

The two families are mirror images. The second family is the derivative table of inverse trig functions read backwards, and the first family is produced by integrating by parts until an integral from the second family (or a simple algebraic one) appears.

The functions and their principal ranges, which every row in this list assumes:

These are the principal values used by most textbooks and calculators. A few older texts choose a different range for arcsec and arccot, which changes the sign pattern of their rows, so check a source's convention before copying from it.

How Do You Read A Table Of Inverse Trig Integrals?

Each row says "the derivative of the result equals the integrand, inside the stated domain". Three reading rules matter more here than in most tables.

• $\sin^{-1}x$ is not $\frac{1}{\sin x}$. The $-1$ marks an inverse function, not a power; the reciprocal is $\csc x$, which has a completely different integral.

• Domains are narrow. $\arcsin x$ and $\arccos x$ exist only on $[-1, 1]$, and $\operatorname{arcsec} x$ only for $\lvert x\rvert \geq 1$, so many rows carry a condition.

• Different answers can match. Since $\arcsin x + \arccos x = \frac{\pi}{2}$, an answer written with $\arcsin$ and one written with $-\arccos$ can differ by a constant and both be correct.

What Are The Integrals Of The Six Inverse Trigonometric Functions?

The six integrals below all come from integration by parts with $u$ equal to the inverse trig function and $dv = dx$. The inverse function becomes its algebraic derivative, and what remains is a square-root or logarithm integral.

The arcsec and arccsc rows split by sign because the derivative of $\operatorname{arcsec} x$ is $\frac{1}{\lvert x\rvert\sqrt{x^2-1}}$, with an absolute value. Many short tables print only the $x > 1$ row, which gives a wrong sign on the left branch.

A Quora user asks, "What is the integral of $\int \arcsin(x),dx$?" Set $u = \arcsin x$ and $dv = dx$, so $du = \frac{dx}{\sqrt{1-x^2}}$ and $v = x$:

$$\int \arcsin x,dx = x\arcsin x - \int \frac{x}{\sqrt{1-x^2}},dx$$

$$= x\arcsin x + \sqrt{1-x^2} + C$$

The last integral uses $w = 1 - x^2$, $dw = -2x,dx$, which turns it into $-\frac{1}{2}\int w^{-1/2},dw = -\sqrt{1-x^2}$.

What Changes When The Argument Is Ax Or A Fraction?

With a scaled argument the results keep their shape and pick up factors from the chain rule. The rows below hold for any constant $a \neq 0$ (and $a > 0$ where $a$ sits under a root or in a denominator as a length).

What Are The Integrals Of Powers Of X Times Inverse Trig Functions?

When a power of $x$ multiplies an inverse trig function, parts still takes $u$ as the inverse function, and $dv = x^n,dx$. The leftover integral is rational or contains $\sqrt{1-x^2}$.

The last three rows are substitutions, not parts: the factor beside the inverse function is its own derivative, so $u = \arctan x$ or $u = \arcsin x$ finishes them in one step.

A homework.study.com question asks to "Evaluate the integral $\int x\arcsin x,dx$." Parts with $u = \arcsin x$ and $dv = x,dx$ gives

$$\int x\arcsin x,dx = \frac{x^2}{2}\arcsin x - \frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}},dx$$

The remaining integral needs $x = \sin\theta$, which gives $\int \frac{x^2}{\sqrt{1-x^2}},dx = \frac{1}{2}\arcsin x - \frac{x\sqrt{1-x^2}}{2}$. Substituting back and collecting the arcsin terms gives the first row of the table, $\frac{2x^2-1}{4}\arcsin x + \frac{x\sqrt{1-x^2}}{4} + C$.

Which Integrals Produce Inverse Trigonometric Functions?

Three algebraic shapes integrate to inverse trig functions, and every inverse-trig-producing integral in a first course is one of them after a substitution. They are the derivatives of inverse trigonometric functions read backwards, with a constant $a > 0$ built in.

Notice that arccos, arccot and arccsc never need their own "producing" rows: their derivatives are the negatives of the arcsin, arctan and arcsec derivatives, so the same integrals cover them up to a constant.

To see where the arcsin row comes from, substitute $x = a\sin\theta$, which is trigonometric substitution in its simplest form. Then $dx = a\cos\theta,d\theta$ and $\sqrt{a^2 - x^2} = a\cos\theta$ on the principal range, so

$$\int \frac{dx}{\sqrt{a^2-x^2}} = \int \frac{a\cos\theta}{a\cos\theta},d\theta = \theta + C = \arcsin\frac{x}{a} + C$$

The arcsec row is the one with a trap. A Brainly question asks to "Use the indicated substitution to evaluate the integral $\int \frac{1}{x\sqrt{x^2-361}},dx$". Here $a = 19$, so the answer is $\frac{1}{19}\operatorname{arcsec}\frac{\lvert x\rvert}{19} + C$. Drop the absolute value and the formula gives the wrong sign for every $x < -19$, because $\frac{1}{x\sqrt{x^2-361}}$ is negative there while $\operatorname{arcsec}$ is increasing.

How Does Completing The Square Lead To An Inverse Trig Integral?

A quadratic that does not factor over the reals can be rewritten as "something squared plus a constant", and then it matches an arctan or arcsin row. Completing the square is the bridge.

For a denominator $ax^2 + bx + c$ with $b^2 < 4ac$ (no real roots), the general result is

$$\int \frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\arctan\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right) + C$$

Two worked rewrites show the pattern:

$$x^2 + 2x + 5 = (x+1)^2 + 4 \quad\Rightarrow\quad \int \frac{dx}{x^2+2x+5} = \frac{1}{2}\arctan\frac{x+1}{2} + C$$

$$3 + 2x - x^2 = 4 - (x-1)^2 \quad\Rightarrow\quad \int \frac{dx}{\sqrt{3+2x-x^2}} = \arcsin\frac{x-1}{2} + C$$

When the quadratic does have real roots ($b^2 > 4ac$), the same integral splits into partial fractions and the answer is a logarithm, not an arctangent. The sign of $b^2 - 4ac$ decides the family before any algebra starts.

Which Inverse Trig Integrals Have No Elementary Formula?

Some integrands that look one step harder than the table have no antiderivative made of elementary functions. A Quora user asks, "What is the integration of tan inverse X divided by X?" The honest answer is that $\int \frac{\arctan x}{x},dx$ has no elementary formula; it defines a special function, the inverse tangent integral,

$$\operatorname{Ti}2(x) = \int{0}^{x} \frac{\arctan t}{t},dt = x - \frac{x^3}{3^2} + \frac{x^5}{5^2} - \frac{x^7}{7^2} + \cdots$$

Its value at $1$ is Catalan's constant, $G = \operatorname{Ti}2(1) \approx 0.9160$, a number whose irrationality is still unproved. The neighbouring integral$\int \frac{\arcsin x}{x},dx$is also non-elementary, yet its definite value is clean:$\int{0}^{1} \frac{\arcsin x}{x},dx = \frac{\pi}{2}\ln 2 \approx 1.0888$. The series and the link to Catalan's constant are set out on the inverse tangent integral reference page.

What Are The Standard Definite Integrals In This List?

Many definite values in this family are exact multiples of $\pi$, because the antiderivative is an angle. The definite integral is just the difference of two angles.

The $\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}$ and $\int_{-\infty}^{\infty}$ rows are improper integrals: one integrand blows up at $x = 1$, the other interval is infinite, and both converge. The row $\int_{0}^{1} \sqrt{1-x^2},dx = \frac{\pi}{4}$ is the area of a quarter of the unit circle, which is exactly where the arcsin term in its antiderivative comes from.

Examples Of Integrals Of Inverse Trigonometric Functions

Example 1

Use the list to find $\int \frac{dx}{4 + 9x^2}$.

Factor out the $9$ so the $x^2$ term has coefficient $1$:

$$\int \frac{dx}{4 + 9x^2} = \frac{1}{9}\int \frac{dx}{\frac{4}{9} + x^2}$$

Use the arctan row with $a = \frac{2}{3}$:

$$= \frac{1}{9}\cdot\frac{3}{2}\arctan\frac{3x}{2} + C = \frac{1}{6}\arctan\frac{3x}{2} + C$$

Final answer: $\frac{1}{6}\arctan\frac{3x}{2} + C$

Example 2

Find $\int \frac{dx}{4 + x^2}$.

The shape $\frac{1}{1+x^2}$ integrates to $\arctan x$, so it is tempting to scale the inside only:

$$\text{Attempt: } \int \frac{dx}{4 + x^2} = \arctan\frac{x}{2} + C$$

Differentiate the attempt to check it:

$$\frac{d}{dx}\arctan\frac{x}{2} = \frac{\frac{1}{2}}{1 + \frac{x^2}{4}} = \frac{2}{4 + x^2}$$

That is twice the integrand, so the attempt is off by a factor of $2$. The row $\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\frac{x}{a} + C$ carries a $\frac{1}{a}$ in front, and here $a = 2$:

$$\int \frac{dx}{4 + x^2} = \frac{1}{2}\arctan\frac{x}{2} + C$$

Final answer: $\frac{1}{2}\arctan\frac{x}{2} + C$

Example 3

Find $\int \frac{2x + 3}{x^2 + 2x + 5},dx$.

Split the numerator into a multiple of the denominator's derivative, $2x + 2$, plus a constant:

$$\frac{2x+3}{x^2+2x+5} = \frac{2x+2}{x^2+2x+5} + \frac{1}{x^2+2x+5}$$

The first part is $\frac{f'(x)}{f(x)}$, and the second completes the square to $(x+1)^2 + 4$:

$$\int \frac{2x+2}{x^2+2x+5},dx = \ln\big(x^2+2x+5\big)$$

$$\int \frac{dx}{(x+1)^2+4} = \frac{1}{2}\arctan\frac{x+1}{2}$$

No absolute value is needed in the logarithm, since $x^2 + 2x + 5 > 0$ for every real $x$.

Final answer: $\ln\big(x^2+2x+5\big) + \frac{1}{2}\arctan\frac{x+1}{2} + C$

Example 4

Evaluate $\int_{0}^{1} \arctan x,dx$.

Use the arctan row and evaluate at the limits:

$$\int_{0}^{1} \arctan x,dx = \left[x\arctan x - \frac{1}{2}\ln\big(1+x^2\big)\right]_{0}^{1}$$

$$= \left(\frac{\pi}{4} - \frac{1}{2}\ln 2\right) - (0 - 0)$$

Final answer: $\frac{\pi}{4} - \frac{\ln 2}{2} \approx 0.4388$

Example 5

Find $\int \frac{dx}{\sqrt{7 + 6x - x^2}}$.

Complete the square under the root:

$$7 + 6x - x^2 = 16 - \big(x^2 - 6x + 9\big) = 16 - (x-3)^2$$

Now it matches the arcsin row with $a = 4$ and the inside $x - 3$:

$$\int \frac{dx}{\sqrt{16 - (x-3)^2}} = \arcsin\frac{x-3}{4} + C$$

The domain is $\lvert x - 3\rvert < 4$, that is $-1 < x < 7$. The step most often skipped here is pulling the minus sign out of $-x^2 + 6x$ before completing the square; leave it in and the constant comes out as $-2$ instead of $16$.

Final answer: $\arcsin\frac{x-3}{4} + C$

Example 6

A lighthouse stands 1 km off a straight coast and flashes its beam in a random direction towards the shore. The point where the beam hits has density $\frac{1}{\pi(1+x^2)}$, with $x$ in km measured along the coast from the nearest point. Find the probability that the beam lands within 1 km of that nearest point.

The probability is the area under the density from $-1$ to $1$:

$$P = \int_{-1}^{1} \frac{dx}{\pi(1+x^2)} = \frac{1}{\pi}\Big[\arctan x\Big]_{-1}^{1}$$

$$= \frac{1}{\pi}\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = \frac{1}{2}$$

Final answer: $\frac{1}{2}$. Half of all flashes land within 1 km, even though the beam can hit points arbitrarily far down the coast.

Why Do Inverse Trig Integrals Matter?

Inverse trig integrals matter because they turn areas into angles, and angles are what geometry, rotation and waves are measured in. Three places make that concrete.

• Pi from algebra. $\int_{0}^{1} \frac{dx}{1+x^2} = \frac{\pi}{4}$ expresses $\pi$ through a rational function. Expanding $\frac{1}{1+x^2}$ as $1 - x^2 + x^4 - \cdots$ and integrating term by term gives $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \cdots$, the series for pi that Madhava of Sangamagrama found and that Gregory (1671) and Leibniz (1673) later rediscovered.

• Circle areas. $\int \sqrt{a^2 - x^2},dx$ contains $\arcsin\frac{x}{a}$ because a slice of a circle's area is partly a sector, and a sector's area depends on its angle.

• Angles from positions. Any time a quantity is the angle seen from a known distance, such as a camera tracking a runner or a beam sweeping a coast in Example 6, an arctangent appears, and its integral or derivative follows.

Where Integrals Of Inverse Trig Functions Go Sideways

Mistake 1: Reading The Inverse As A Reciprocal

Where it slips in: Integrands written as $\tan^{-1}x$ or $\sin^{-1}x$.

Don't do this: Treating $\int \sin^{-1}x,dx$ as $\int \csc x,dx$ and writing $\ln\left\lvert\tan\frac{x}{2}\right\rvert + C$.

The correct way: Read $\sin^{-1}x$ as $\arcsin x$ and use parts: $x\arcsin x + \sqrt{1-x^2} + C$.

The swap between the reciprocal $\frac{1}{\sin x}$ and the inverse $\sin^{-1}x$ is the single most common source of wrong answers in this topic, and the memorizer is most exposed, because "$-1$ means one over" is a rule they learned for numbers and powers. Writing $\arcsin$ instead of $\sin^{-1}$ in your own working removes the ambiguity entirely.

Mistake 2: Using The Right-Branch Arcsec Formula For Negative X

Where it slips in: Integrals such as $\int \frac{dx}{x\sqrt{x^2-a^2}}$ or $\int \operatorname{arcsec} x,dx$ evaluated for $x < -a$ or $x < -1$.

Don't do this: Writing $\frac{1}{a}\operatorname{arcsec}\frac{x}{a}$ without the absolute value and using it for negative $x$.

The correct way: Use $\frac{1}{a}\operatorname{arcsec}\frac{\lvert x\rvert}{a} + C$, and use the $x < -1$ row of the arcsec table on the left branch.

Mistake 3: Losing The Leading Coefficient After Completing The Square

Where it slips in: Denominators like $4x^2 + 4x + 5$, where the $x^2$ coefficient is not $1$.

Don't do this: Rewriting $4x^2 + 4x + 5$ as $\big(x + \frac{1}{2}\big)^2 + 1$ and answering $\arctan\big(x + \frac{1}{2}\big) + C$.

The correct way: Keep the $4$: $4x^2 + 4x + 5 = 4\left[\big(x + \frac{1}{2}\big)^2 + 1\right]$, so the answer is $\frac{1}{4}\arctan\big(x + \frac{1}{2}\big) + C$.

The rusher completes the square on autopilot and forgets the factor sitting outside the bracket. Differentiating the final answer once catches it immediately.

Mistake 4: Erasing A Correct Answer Because It Uses Arccos

Where it slips in: Comparing your result with a table or answer key.

Don't do this: Deciding that $-\arccos\frac{x}{a} + C$ is wrong because the key shows $\arcsin\frac{x}{a} + C$.

The correct way: Check whether the two answers differ by a constant. Here $\arcsin\frac{x}{a} = \frac{\pi}{2} - \arccos\frac{x}{a}$, so both are valid antiderivatives.

The second-guesser loses more marks to this than to any real error, crossing out a correct line under time pressure.

The real-world version of Mistake 2 lives in software. The one-argument arctangent returns angles only between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, so a program that converts a position to a heading with plain arctan silently reports the wrong direction whenever the point lies to the left of the origin. That branch problem is why almost every programming language ships a separate two-argument function, atan2, which uses the signs of both coordinates to pick the right quadrant.

Practice Problems On The List Of Integrals Of Inverse Trigonometric Functions

1. Find $\int \frac{dx}{\sqrt{25 - x^2}}$.

2. Find $\int \frac{dx}{x^2 + 6x + 13}$.

3. Find $\int \arccos(2x),dx$.

4. Find $\int \frac{x}{1 + x^4},dx$.

5. Evaluate $\int_{0}^{1} x\arctan x,dx$.

Answer to Question 1: $\arcsin\frac{x}{5} + C$

Answer to Question 2: $\frac{1}{2}\arctan\frac{x+3}{2} + C$

Answer to Question 3: $x\arccos(2x) - \frac{\sqrt{1-4x^2}}{2} + C$

Answer to Question 4: $\frac{1}{2}\arctan\big(x^2\big) + C$

Answer to Question 5: $\frac{\pi}{4} - \frac{1}{2} \approx 0.2854$

Wrapping Up

• The list of integrals of inverse trigonometric functions has two families: integrals of arcsin, arctan and their relatives, and integrals that produce them.

• The first family comes from integration by parts with $u$ as the inverse function and $dv = dx$.

• The second family is $\frac{1}{\sqrt{a^2-x^2}}$, $\frac{1}{a^2+x^2}$ and $\frac{1}{x\sqrt{x^2-a^2}}$, each with its own factor of $\frac{1}{a}$ or none.

• Completing the square turns many quadratic denominators into an arctan or arcsin row, but only when the quadratic has no real roots.

• Arcsec rows need $\lvert x\rvert$, and $\int \frac{\arctan x}{x},dx$ has no elementary formula.

A Practical Next Step

Work through practice problems 1 to 5; if Problem 2 or 3 trips you up, return to the completing-the-square section or the scaled-argument table and match the row before integrating. Want a live Bhanzu trainer to practise these recognitions with you? Book a free demo class.

Read More

• List of integrals for every function family in one reference

• Table of integrals for the power, exponential and trig rows these results sit beside

• Basic integration formulas that every row here builds on

• Integration by substitution for the inside-function moves used in the scaled rows

• Integration of rational functions for the cases where a quadratic denominator gives a logarithm instead

• Inverse functions and differentiation for the derivative rules this list reverses