# 5.7 Integrals resulting in inverse trigonometric functions

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• 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

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

find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one
state and prove L hospital rule
I want to know about hospital rule
Faysal
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were$48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach \$60 billion?
what is the derivative of x= Arc sin (x)^1/2
y^2 = arcsin(x)
Pitior
x = sin (y^2)
Pitior
differentiate implicitly
Pitior
then solve for dy/dx
Pitior
thank you it was very helpful
morfling
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.