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Integrals Resulting in Inverse Trigonometric Functions

Integrate functions resulting in inverse trigonometric functions

Integrals that Result in Inverse Sine Functions

Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

Condensed — the full section is in OpenStax Calculus Volume 2.

Integrals Resulting in Other Inverse Trigonometric Functions

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.

Example

Try it.

Evaluate the integral \(\int \frac{1}{1+4{x}^{2}}dx.\)

Solution

Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for \({\text{tan}}^{-1}u+C.\) So we use substitution, letting \(u=2x,\) then \(du=2dx\) and \(1\text{/}2du=dx.\) Then, we have

\[\frac{1}{2}\int \frac{1}{1+{u}^{2}}du=\frac{1}{2}\ {\text{tan}}^{-1}u+C=\frac{1}{2}\ {\text{tan}}^{-1}(2x)+C.\]

Example

Try it.

Evaluate the integral \(\int \frac{1}{9+{x}^{2}}dx.\)

Solution

Apply the formula with \(a=3.\) Then,

\[\int \frac{dx}{9+{x}^{2}}=\frac{1}{3}\ {\text{tan}}^{-1}(\frac{x}{3})+C.\]

Example

Try it.

Evaluate the definite integral \({\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}.\)

Solution

Use the formula for the inverse tangent. We have

\[\begin{array}{ll} \\ \\ {\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}} & ={\text{tan}}^{-1}x{|}_{\sqrt{3}\text{/}3}^{\sqrt{3}} \\ & =[{\text{tan}}^{-1}(\sqrt{3})]-[{\text{tan}}^{-1}(\frac{\sqrt{3}}{3})] \\ & =\frac{\pi }{6}.\end{array}\]

Key Concepts

Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions.
Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
Substitution is often required to put the integrand in the correct form.

Key Equations

Integrals That Produce Inverse Trigonometric Functions

\(\int \frac{du}{\sqrt{{a}^{2}-{u}^{2}}}={\text{sin}}^{-1}(\frac{u}{a})+C\)
\(\int \frac{du}{{a}^{2}+{u}^{2}}=\frac{1}{a}\ {\text{tan}}^{-1}(\frac{u}{a})+C\)
\(\int \frac{du}{u\sqrt{{u}^{2}-{a}^{2}}}=\frac{1}{a}\ {\text{sec}}^{-1}(\frac{u}{a})+C\)

Integrals Resulting in Inverse Trigonometric Functions

In the following exercises, evaluate each integral in terms of an inverse trigonometric function.

In the following exercises, find each indefinite integral, using appropriate substitutions.

In the following exercises, solve for the antiderivative \(\int f\) of f with \(C=0,\) then use a calculator to graph f and the antiderivative over the given interval \([a,b].\) Identify a value of C such that adding C to the antiderivative recovers the definite integral \(F(x)={\int }_{a}^{x}f(t)dt.\)

In the following exercises, compute the antiderivative using appropriate substitutions.

In the following exercises, use a calculator to graph the antiderivative \(\int f\) with \(C=0\) over the given interval \([a,b].\) Approximate a value of C, if possible, such that adding C to the antiderivative gives the same value as the definite integral \(F(x)={\int }_{a}^{x}f(t)dt.\)

In the following exercises, compute each integral using appropriate substitutions.

In the following exercises, compute each definite integral.

Integrals that Result in Inverse Sine Functions

Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

Condensed — the full section is in OpenStax Calculus Volume 1.

Integrals Resulting in Other Inverse Trigonometric Functions

Example

Try it.

Evaluate the integral \(\int \frac{1}{1+4{x}^{2}}dx.\)

Solution

\[\frac{1}{2}\int \frac{1}{1+{u}^{2}}du=\frac{1}{2}\ {\text{tan}}^{-1}u+C=\frac{1}{2}\ {\text{tan}}^{-1}(2x)+C.\]

Example

Try it.

Evaluate the integral \(\int \frac{1}{9+{x}^{2}}dx.\)

Solution

Apply the formula with \(a=3.\) Then,

\[\int \frac{dx}{9+{x}^{2}}=\frac{1}{3}\ {\text{tan}}^{-1}(\frac{x}{3})+C.\]

Example

Try it.

Evaluate the definite integral \({\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}.\)

Solution

Use the formula for the inverse tangent. We have

Key Concepts

Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions.
Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
Substitution is often required to put the integrand in the correct form.

Key Equations

Integrals That Produce Inverse Trigonometric Functions

Integrals Resulting in Inverse Trigonometric Functions

In the following exercises, evaluate each integral in terms of an inverse trigonometric function.

In the following exercises, find each indefinite integral, using appropriate substitutions.

In the following exercises, compute the antiderivative using appropriate substitutions.

In the following exercises, compute each integral using appropriate substitutions.

In the following exercises, compute each definite integral.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Evaluate the definite integral \({\int }_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-{x}^{2}}}.\)

Révèle la réponse

We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have
\[\begin{array}{ll} \\ \\ {\int }_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-{x}^{2}}} & ={\text{sin}}^{-1}x{|}_{0}^{\frac{1}{2}} \\ & ={\text{sin}}^{-1}\frac{1}{2}-{\text{sin}}^{-1}0 \\ & =\frac{\pi }{6}-0 \\ & =\frac{\pi }{6}.\end{array}\]
Evaluate the integral \(\int \frac{dx}{\sqrt{1-16{x}^{2}}}.\)

Révèle la réponse

\(\frac{1}{4}\ {\text{sin}}^{-1}(4x)+C\)
Evaluate the integral \(\int \frac{dx}{\sqrt{4-9{x}^{2}}}.\)

Révèle la réponse

Substitute \(u=3x.\) Then \(du=3dx\) and we have
\[\int \frac{dx}{\sqrt{4-9{x}^{2}}}=\frac{1}{3}\int \frac{du}{\sqrt{4-{u}^{2}}}.\]
Applying the formula with \(a=2,\) we obtain
\[\begin{array}{ll}\int \frac{dx}{\sqrt{4-9{x}^{2}}} & =\frac{1}{3}\int \frac{du}{\sqrt{4-{u}^{2}}} \\ \\ & =\frac{1}{3}{\text{sin}}^{-1}(\frac{u}{2})+C \\ & =\frac{1}{3}{\text{sin}}^{-1}(\frac{3x}{2})+C.\end{array}\]
Find the indefinite integral using an inverse trigonometric function and substitution for \(\int \frac{dx}{\sqrt{9-{x}^{2}}}.\)

Révèle la réponse

\({\text{sin}}^{-1}(\frac{x}{3})+C\)
Evaluate the definite integral \({\int }_{0}^{\sqrt{3}\text{/}2}\frac{du}{\sqrt{1-{u}^{2}}}.\)

Révèle la réponse

The format of the problem matches the inverse sine formula. Thus,
\[\begin{array}{ll} \\ \\ {\int }_{0}^{\sqrt{3}\text{/}2}\frac{du}{\sqrt{1-{u}^{2}}} & ={\text{sin}}^{-1}u{|}_{0}^{\sqrt{3}\text{/}2} \\ & =[{\text{sin}}^{-1}(\frac{\sqrt{3}}{2})]-[{\text{sin}}^{-1}(0)] \\ & =\frac{\pi }{3}.\end{array}\]
Evaluate the integral \(\int \frac{1}{1+4{x}^{2}}dx.\)

Révèle la réponse

Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for \({\text{tan}}^{-1}u+C.\) So we use substitution, letting \(u=2x,\) then \(du=2dx\) and \(1\text{/}2du=dx.\) Then, we have
\[\frac{1}{2}\int \frac{1}{1+{u}^{2}}du=\frac{1}{2}\ {\text{tan}}^{-1}u+C=\frac{1}{2}\ {\text{tan}}^{-1}(2x)+C.\]
Use substitution to find the antiderivative \(\int \frac{dx}{25+4{x}^{2}}.\)

Révèle la réponse

\(\frac{1}{10}\ {\text{tan}}^{-1}(\frac{2x}{5})+C\)
Evaluate the integral \(\int \frac{1}{9+{x}^{2}}dx.\)

Révèle la réponse

Apply the formula with \(a=3.\) Then,
\[\int \frac{dx}{9+{x}^{2}}=\frac{1}{3}\ {\text{tan}}^{-1}(\frac{x}{3})+C.\]
Evaluate the integral \(\int \frac{dx}{16+{x}^{2}}.\)

Révèle la réponse

\(\frac{1}{4}\ {\text{tan}}^{-1}(\frac{x}{4})+C\)
Evaluate the definite integral \({\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}.\)

Révèle la réponse

Use the formula for the inverse tangent. We have
\[\begin{array}{ll} \\ \\ {\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}} & ={\text{tan}}^{-1}x{|}_{\sqrt{3}\text{/}3}^{\sqrt{3}} \\ & =[{\text{tan}}^{-1}(\sqrt{3})]-[{\text{tan}}^{-1}(\frac{\sqrt{3}}{3})] \\ & =\frac{\pi }{6}.\end{array}\]
Evaluate the definite integral \({\int }_{0}^{2}\frac{dx}{4+{x}^{2}}.\)

Révèle la réponse

\(\frac{\pi }{8}\)
\({\int }_{0}^{\sqrt{3}\text{/}2}\frac{dx}{\sqrt{1-{x}^{2}}}\)

Révèle la réponse

\({\text{sin}}^{-1}x{|}_{0}^{\sqrt{3}\text{/}2}=\frac{\pi }{3}\)
\({\int }_{-1\text{/}2}^{1\text{/}2}\frac{dx}{\sqrt{1-{x}^{2}}}\)
\({\int }_{\sqrt{3}}^{1}\frac{dx}{1+{x}^{2}}\)

Révèle la réponse

\({\text{tan}}^{-1}x{|}_{\sqrt{3}}^{1}=-\frac{\pi }{12}\)
\({\int }_{1\text{/}\sqrt{3}}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}\)
\({\int }_{\frac{2}{\sqrt{3}}}^{\sqrt{2}}\frac{dx}{|x|\sqrt{{x}^{2}-1}}\)

Révèle la réponse

\({\text{sec}}^{-1}{|}_{\frac{2}{\sqrt{3}}}^{\sqrt{2}}=\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}\)
\({\int }_{\sqrt{2}}^{2}\frac{dx}{|x|\sqrt{{x}^{2}-1}}\)
\(\int \frac{dx}{\sqrt{9-{x}^{2}}}\)

Révèle la réponse

\({\text{sin}}^{-1}(\frac{x}{3})+C\)
\(\int \frac{dx}{\sqrt{1-16{x}^{2}}}\)
\(\int \frac{dx}{9+{x}^{2}}\)

Révèle la réponse

\(\frac{1}{3}\ {\text{tan}}^{-1}(\frac{x}{3})+C\)
\(\int \frac{dx}{25+16{x}^{2}}\)
\(\int \frac{dx}{|x|\sqrt{{x}^{2}-9}}\)

Révèle la réponse

\(\frac{1}{3}\ {\text{sec}}^{-1}(\frac{x}{3})+C\)
\(\int \frac{dx}{|x|\sqrt{4{x}^{2}-16}}\)
Explain the relationship \(\text{-}{\text{cos}}^{-1}t+C=\int \frac{dt}{\sqrt{1-{t}^{2}}}={\text{sin}}^{-1}t+C.\) Is it true, in general, that \({\text{cos}}^{-1}t=\text{-}{\text{sin}}^{-1}t?\)

Révèle la réponse

\(\text{cos}(\frac{\pi }{2}-\theta )=\text{sin}\ \theta .\) So, \({\text{sin}}^{-1}t=\frac{\pi }{2}-{\text{cos}}^{-1}t.\) They differ by a constant.
Explain the relationship \({\text{sec}}^{-1}t+C=\int \frac{dt}{|t|\sqrt{{t}^{2}-1}}=\text{-}{\text{csc}}^{-1}t+C.\) Is it true, in general, that \({\text{sec}}^{-1}t=\text{-}{\text{csc}}^{-1}t?\)
Explain what is wrong with the following integral: \({\int }_{1}^{2}\frac{dt}{\sqrt{1-{t}^{2}}}.\)

Révèle la réponse

\(\sqrt{1-{t}^{2}}\) is not defined as a real number when \(t>1.\)
Explain what is wrong with the following integral: \({\int }_{-1}^{1}\frac{dt}{|t|\sqrt{{t}^{2}-1}}.\)
[T] \(\int \frac{1}{\sqrt{9-{x}^{2}}}dx\) over \([-3,3]\)

Révèle la réponse

The antiderivative is \({\text{sin}}^{-1}(\frac{x}{3})+C.\) Taking \(C=\frac{\pi }{2}\) recovers the definite integral.
[T] \(\int \frac{9}{9+{x}^{2}}dx\) over \([-6,6]\)
[T] \(\int \frac{\text{cos}\ x}{4+{\text{sin}}^{2}x}dx\) over \([-6,6]\)

Révèle la réponse

The antiderivative is \(\frac{1}{2}\ {\text{tan}}^{-1}(\frac{\text{sin}\ x}{2})+C.\) Taking \(C=\frac{1}{2}\ {\text{tan}}^{-1}(\frac{\text{sin}(6)}{2})\) recovers the definite integral.
[T] \(\int \frac{{e}^{x}}{1+{e}^{2x}}dx\) over \([-6,6]\)
\(\int \frac{{\text{sin}}^{-1}tdt}{\sqrt{1-{t}^{2}}}\)

Révèle la réponse

\(\frac{1}{2}{({\text{sin}}^{-1}t)}^{2}+C\)
\(\int \frac{dt}{{\text{sin}}^{-1}t\sqrt{1-{t}^{2}}}\)
\(\int \frac{{\text{tan}}^{-1}(2t)}{1+4{t}^{2}}dt\)

Révèle la réponse

\(\frac{1}{4}{({\text{tan}}^{-1}(2t))}^{2}+C\)
\(\int \frac{t{\text{tan}}^{-1}({t}^{2})}{1+{t}^{4}}dt\)
\(\int \frac{{\text{sec}}^{-1}(\frac{t}{2})}{|t|\sqrt{{t}^{2}-4}}dt\)

Révèle la réponse

\(\frac{1}{4}({\text{sec}}^{-1}{(\frac{t}{2})}^{2})+C\)
\(\int \frac{t{\text{sec}}^{-1}({t}^{2})}{{t}^{2}\sqrt{{t}^{4}-1}}dt\)
[T] \(\int \frac{1}{x\sqrt{{x}^{2}-4}}dx\) over \([2,6]\)

Révèle la réponse

The antiderivative is \(\frac{1}{2}\ {\text{sec}}^{-1}(\frac{x}{2})+C.\) Taking \(C=0\) recovers the definite integral over \([2,6].\)
[T] \(\int \frac{1}{(2x+2)\sqrt{x}}dx\) over \([0,6]\)
[T] \(\int \frac{(\text{sin}\ x+x\ \text{cos}\ x)}{1+{x}^{2}{\text{sin}}^{2}x}dx\) over \([-6,6]\)

Révèle la réponse

The general antiderivative is \({\text{tan}}^{-1}(x\ \text{sin}\ x)+C.\) Taking \(C=\text{-}{\text{tan}}^{-1}(6\ \text{sin}(6))\) recovers the definite integral.

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Integrals Resulting in Inverse Trigonometric Functions

Integrate functions resulting in inverse trigonometric functions
Formulas for derivatives of inverse trigonometric functions developed in
Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
Substitution is often required to put the integrand in the correct form.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests