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Use substitution to find the antiderivative of d x 25 + 4 x 2 .

1 10 tan −1 ( 2 x 5 ) + C

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Applying the integration formulas

Find the antiderivative of 1 9 + x 2 d x .

Apply the formula with a = 3 . Then,

d x 9 + x 2 = 1 3 tan −1 ( x 3 ) + C .
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Find the antiderivative of d x 16 + x 2 .

1 4 tan −1 ( x 4 ) + C

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Evaluating a definite integral

Evaluate the definite integral 3 / 3 3 d x 1 + x 2 .

Use the formula for the inverse tangent. We have

3 / 3 3 d x 1 + x 2 = tan −1 x | 3 / 3 3 = [ tan −1 ( 3 ) ] [ tan −1 ( 3 3 ) ] = π 6 .
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Evaluate the definite integral 0 2 d x 4 + x 2 .

π 8

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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
    d u a 2 u 2 = sin −1 ( u a ) + C
    d u a 2 + u 2 = 1 a tan −1 ( u a ) + C
    d u u u 2 a 2 = 1 a sec −1 ( u a ) + C

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

0 3 / 2 d x 1 x 2

sin −1 x | 0 3 / 2 = π 3

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−1 / 2 1 / 2 d x 1 x 2

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3 1 d x 1 + x 2

tan −1 x | 3 1 = π 12

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1 2 d x | x | x 2 1

sec −1 x | 1 2 = π 4

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1 2 / 3 d x | x | x 2 1

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

d x 9 x 2

sin −1 ( x 3 ) + C

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d x 9 + x 2

1 3 tan −1 ( x 3 ) + C

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d x | x | x 2 9

1 3 sec −1 ( x 3 ) + C

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d x | x | 4 x 2 16

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Explain the relationship cos −1 t + C = d t 1 t 2 = sin −1 t + C . Is it true, in general, that cos −1 t = sin −1 t ?

cos ( π 2 θ ) = sin θ . So, sin −1 t = π 2 cos −1 t . They differ by a constant.

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Explain the relationship sec −1 t + C = d t | t | t 2 1 = csc −1 t + C . Is it true, in general, that sec −1 t = csc −1 t ?

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Explain what is wrong with the following integral: 1 2 d t 1 t 2 .

1 t 2 is not defined as a real number when t > 1 .

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Explain what is wrong with the following integral: −1 1 d t | t | t 2 1 .

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In the following exercises, solve for the antiderivative 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 ) = a x f ( t ) d t .

[T] 1 9 x 2 d x over [ −3 , 3 ]


Two graphs. The first shows the function f(x) = 1 / sqrt(9 – x^2). It is an upward opening curve symmetric about the y axis, crossing at (0, 1/3). The second shows the function F(x) = arcsin(1/3 x). It is an increasing curve going through the origin.
The antiderivative is sin −1 ( x 3 ) + C . Taking C = π 2 recovers the definite integral.

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[T] 9 9 + x 2 d x over [ −6 , 6 ]

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[T] cos x 4 + sin 2 x d x over [ −6 , 6 ]


Two graphs. The first shows the function f(x) = cos(x) / (4 + sin(x)^2). It is an oscillating function over [-6, 6] with turning points at roughly (-3, -2.5), (0, .25), and (3, -2.5), where (0,.25) is a local max and the others are local mins. The second shows the function F(x) = .5 * arctan(.5*sin(x)), which also oscillates over [-6,6]. It has turning points at roughly (-4.5, .25), (-1.5, -.25), (1.5, .25), and (4.5, -.25).
The antiderivative is 1 2 tan −1 ( sin x 2 ) + C . Taking C = 1 2 tan −1 ( sin ( 6 ) 2 ) recovers the definite integral.

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[T] e x 1 + e 2 x d x over [ −6 , 6 ]

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

sin −1 t d t 1 t 2

1 2 ( sin −1 t ) 2 + C

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d t sin −1 t 1 t 2

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tan −1 ( 2 t ) 1 + 4 t 2 d t

1 4 ( tan −1 ( 2 t ) ) 2

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t tan −1 ( t 2 ) 1 + t 4 d t

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sec −1 ( t 2 ) | t | t 2 4 d t

1 4 ( sec −1 ( t 2 ) 2 ) + C

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t sec −1 ( t 2 ) t 2 t 4 1 d t

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In the following exercises, use a calculator to graph the antiderivative 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 ) = a x f ( t ) d t .

[T] 1 x x 2 4 d x over [ 2 , 6 ]


A graph of the function f(x) = -.5 * arctan(2 / ( sqrt(x^2 – 4) ) ) in quadrant four. It is an increasing concave down curve with a vertical asymptote at x=2.
The antiderivative is 1 2 sec −1 ( x 2 ) + C . Taking C = 0 recovers the definite integral over [ 2 , 6 ] .

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[T] 1 ( 2 x + 2 ) x d x over [ 0 , 6 ]

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[T] ( sin x + x cos x ) 1 + x 2 sin 2 x d x over [ −6 , 6 ]


The graph of f(x) = arctan(x sin(x)) over [-6,6]. It has five turning points at roughly (-5, -1.5), (-2,1), (0,0), (2,1), and (5,-1.5).
The general antiderivative is tan −1 ( x sin x ) + C . Taking C = tan −1 ( 6 sin ( 6 ) ) recovers the definite integral.

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Questions & Answers

Given that u = tan–¹(y/x), show that d²u/dx² + d²u/dy²=0
Collince Reply
find the limiting value of 5n-3÷2n-7
Joy Reply
Use the first principal to solve the following questions 5x-1
Cecilia Reply
175000/9*100-100+164294/9*100-100*4
Ibrahim Reply
mode of (x+4) is equal to 10..graph it how?
Sunny Reply
66
ram
6
ram
6
Cajab
what is domain in calculus
nelson
integrals of 1/6-6x-5x²
Namwandi Reply
derivative of (-x^3+1)%x^2
Misha Reply
(-x^5+x^2)/100
Sarada
(-5x^4+2x)/100
Sarada
oh sorry it's (-x^3+1)÷x^2
Misha
-5x^4+2x
Sarada
sorry I didn't understan A with that symbol
Sarada
find the derivative of the following y=4^e5x y=Cos^2 y=x^inx , x>0 y= 1+x^2/1-x^2 y=Sin ^2 3x + Cos^2 3x please guys I need answer and solutions
Ga Reply
differentiate y=(3x-2)^2(2x^2+5) and simplify the result
Ga
72x³-72x²+106x-60
okhiria
y= (2x^2+5)(3x+9)^2
lemmor
solve for dy/dx of y= 8x^3+5x^2-x+5
Ga Reply
192x^2+50x-1
Daniel
are you sure? my answer is 24x^2+10x-1 but I'm not sure about my answer .. what do you think?
Ga
24x²+10x-1
Eyad
eyad Amin that's the correct answer?
Ga
yes
Eyad
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Ga
hahaha 😂❤️❤️❤️ welcome bro ❤️
Eyad
eyad please answer my other question for my assignment
Ga
y= (2x^2+5)(3x+9)^2
lemmor
can i join?
Fernando
yes of course
Jug
can anyone teach me integral calculus?
Jug
it's just the opposite of differential calculus
yhin
of coursr
okhiria
but i think, it's more complicated than calculus 1
Jug
Hello can someone help me with calculus one...
Jainaba
find the derivative of y= (2x+3)raise to 2 sorry I didn't know how to put the raise correctly
Ga Reply
8x+12
Dhruv
8x+3
okhiria
d the derivative of y= e raised to power x
okhiria
rates of change and tangents to curves
Kyaw Reply
how can find differential Calculus
Kyaw
derivative of ^5√1+x
Rohit Reply
can you help with this f(×)=square roots 3-4
oscar
using first principle, find the derivative of y=cosx^3
RUBY Reply
Approximate root 4 without a calculator
Tinkeu Reply
Approximate root 4.02 without using a calculator
Tinkeu
2.03
temigbe
unit of energy
Safiyo
< or = or >4.00000001
Fufa
2.01
Cajab
+-2
PRATICK
find the integral of tan
Gagan Reply
sec
PRATICK
sec
PRATICK
tan
dipesh

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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