# 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

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

how do we prove the quadratic formular
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
hi vedant can u help me with some assignments
Solomon
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
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?
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