# 1.7 Integrals resulting in inverse trigonometric functions

 Page 1 / 3
• Integrate functions resulting in inverse trigonometric functions

In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall from Functions and Graphs that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also in Derivatives , we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving 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.

## Rule: integration formulas resulting in inverse trigonometric functions

The following integration formulas yield inverse trigonometric functions:

1. $\int \frac{du}{\sqrt{{a}^{2}-{u}^{2}}}={\text{sin}}^{-1}\frac{u}{a}+C$

2. $\int \frac{du}{{a}^{2}+{u}^{2}}=\frac{1}{a}\phantom{\rule{0.05em}{0ex}}{\text{tan}}^{-1}\frac{u}{a}+C$

3. $\int \frac{du}{u\sqrt{{u}^{2}-{a}^{2}}}=\frac{1}{a}\phantom{\rule{0.05em}{0ex}}{\text{sec}}^{-1}\frac{u}{a}+C$

## Proof

Let $y={\text{sin}}^{-1}\frac{x}{a}.$ Then $a\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}y=x.$ Now let’s use implicit differentiation. We obtain

$\begin{array}{ccc}\hfill \frac{d}{dx}\phantom{\rule{0.2em}{0ex}}\left(a\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}y\right)& =\hfill & \frac{d}{dx}\phantom{\rule{0.2em}{0ex}}\left(x\right)\hfill \\ \\ \hfill a\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}& =\hfill & 1\hfill \\ \hfill \frac{dy}{dx}& =\hfill & \frac{1}{a\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}y}.\hfill \end{array}$

For $-\frac{\pi }{2}\le y\le \frac{\pi }{2},\text{cos}\phantom{\rule{0.1em}{0ex}}y\ge 0.$ Thus, applying the Pythagorean identity ${\text{sin}}^{2}y+{\text{cos}}^{2}y=1,$ we have $\text{cos}\phantom{\rule{0.1em}{0ex}}y=\sqrt{1={\text{sin}}^{2}y}.$ This gives

$\begin{array}{cc}\frac{1}{a\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}y}\hfill & =\frac{1}{a\sqrt{1-{\text{sin}}^{2}y}}\hfill \\ \\ & =\frac{1}{\sqrt{{a}^{2}-{a}^{2}{\text{sin}}^{2}y}}\hfill \\ & =\frac{1}{\sqrt{{a}^{2}-{x}^{2}}}.\hfill \end{array}$

Then for $\text{−}a\le x\le a,$ we have

$\int \frac{1}{\sqrt{{a}^{2}-{u}^{2}}}du={\text{sin}}^{-1}\left(\frac{u}{a}\right)+C.$

## Evaluating a definite integral using inverse trigonometric functions

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

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

$\begin{array}{}\\ \\ {\int }_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}\hfill & ={\text{sin}}^{-1}x{|}_{0}^{1}\hfill \\ & ={\text{sin}}^{-1}1-{\text{sin}}^{-1}0\hfill \\ & =\frac{\pi }{2}-0\hfill \\ & =\frac{\pi }{2}.\hfill \end{array}$

Find the antiderivative of $\int \frac{dx}{\sqrt{1-16{x}^{2}}}.$

$\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{\text{sin}}^{-1}\left(4x\right)+C$

## Finding an antiderivative involving an inverse trigonometric function

Evaluate the integral $\int \frac{dx}{\sqrt{4-9{x}^{2}}}.$

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

$\begin{array}{cc}\int \frac{dx}{\sqrt{4-9{x}^{2}}}\hfill & =\frac{1}{3}\int \frac{du}{\sqrt{4-{u}^{2}}}\hfill \\ \\ & =\frac{1}{3}{\text{sin}}^{-1}\left(\frac{u}{2}\right)+C\hfill \\ & =\frac{1}{3}{\text{sin}}^{-1}\left(\frac{3x}{2}\right)+C.\hfill \end{array}$

Find the indefinite integral using an inverse trigonometric function and substitution for $\int \frac{dx}{\sqrt{9-{x}^{2}}}.$

${\text{sin}}^{-1}\left(\frac{x}{3}\right)+C$

## Evaluating a definite integral

Evaluate the definite integral ${\int }_{0}^{\sqrt{3}\text{/}2}\frac{du}{\sqrt{1-{u}^{2}}}.$

The format of the problem matches the inverse sine formula. Thus,

$\begin{array}{}\\ \\ {\int }_{0}^{\sqrt{3}\text{/}2}\frac{du}{\sqrt{1-{u}^{2}}}\hfill & ={\text{sin}}^{-1}u{|}_{0}^{\sqrt{3}\text{/}2}\hfill \\ & =\left[{\text{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)\right]-\left[{\text{sin}}^{-1}\left(0\right)\right]\hfill \\ & =\frac{\pi }{3}.\hfill \end{array}$

## 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.

## Finding an antiderivative involving the inverse tangent function

Find an antiderivative of $\int \frac{1}{1+4{x}^{2}}dx.$

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

$\frac{1}{2}\int \frac{1}{1+{u}^{2}}du=\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{\text{tan}}^{-1}u+C=\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{\text{tan}}^{-1}\left(2x\right)+C.$

#### Questions & Answers

what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
Abdul Reply
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
Abdul

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 2' conversation and receive update notifications?

 By By By By Abishek Devaraj By Eric Crawford