Integration Technique Cheat Sheet 

Integration Technique

When to use this method

The Method Example

U-sub When you have reverse chain rule

( ( )) '( )f u x u x dx

Choose u and compute du so that ( )f u du is integrable.

23 2

3

2

3

4 2 1

1 1212

4 21

124 2

2x dxx

x

u x du x

dudxu

dx

x

Trigonometric Identities

When you are given

( )

or

cos

n

)

si

(

k

k

x dx

x dx

k is even = use the half angle formulas 2

2

1 cos(2 )cos2

1 cos(2 )sin2

( )

( )

x

x

k is odd = “save” one of the functions for the du, and substitute the rest with the identities

2 2

2 2

cos sinsin cos ( )

( ) 1 ( )( ) 1

x xx x

4 2

2

3 2

2

1 cos(2 )( ) ( )2

1 (1 2cos(2 ) cos (2 ))41 1 cos(4 )(1 2co

cos

or

cos ( ) (1 sin (

s(2 ) )4 2

cos() ))

1

dx d

x dx x

u

x

d

d

x

d

d

u

x

Integration by Parts (traditional)

When you have the product of two functions but neither is the derivative of the other

( ) ( )f x g x dx

Choose ( ) and ( )

'( ) ( )

u f x dv g x

du f x v g x dx

udv uv vdu

Choosing u (log, inverse trig, algebraic, trig, exp)

1 1 2 2

1 1 2 2

sin( ) sin( ) cos( )

sin( ) cos( )

cos( ) sin( )

sin( ) sin( ) cos( ) sin( )

2 sin( ) sin( ) cos( )

sin(

x x x

x x

x x

x x x x

x x x

x e dx x e e x dx

u x dv e u x dv e

du x v e du x v e

x e dx x e x e x e dx

x e dx x e x e

) (sin( ) cos( )12

)x xx e dx e x x

Integration Technique Cheat Sheet 

Integration by Parts (tabular)

When one function is a polynomial and the other is NOT a log function

Sign u (der) dv (int) + u dv - 'u v + ''u v

Partial Fractions

When you are integrating a rational function in which the numerator is not the derivative of the denominator

You are given A and B and factor the denominator into

1 2 1 2

2 1

( )( )( ) ( )

Ax B M Nx r x r x r x r

Ax B M x r N x r

Plug in 1 2 and x r x r

2

4 3711 28x dx

x x

2

4 3711 28 7 4

4 37 ( 4) ( 7)4 7 7 3

x Ax x x xx A x

B

B xx B x A

  

 

2

4 37 3 77 411 28

3ln | 7 | 7 ln | 4 |

x dx dx dxx xx x

x x C

 

Trigonometric Substitution

When you have 2

2

2

or

or

a x

x a

x a

Using the trig identities 2 2

2 2

2 2

1 sin ( ) cos( )

( )1 tans

sec ( )ec ta ( )) n( 1

Choose x to be one of the trig functions so that the square root is cancelled.

2

2 2

3

2 7 tan( ) 7sec

49 49 tan (

49 ( )

7) sec )

47sec (

(

)

dx

d

x d

x dx

x

x

7

x

θ

Integration Technique Cheat Sheet 

Tricky Integrals:

1. sec( )x dx

2

2

sec( ) tan( )sec( ) sec( )sec( ) tan( )

sec ( ) sec( ) tan( ) sec( ) tan( )

ln | sec( ) ta

sec( ) tan( ) sec( ) tan( ) se

n( )

)

|

c (

x xx dx x dxx x

x x x dxx x

u x xdu x x

x

C

d x

u

u x

2. 1

x dxx

1 11 1

1 11 1

111

ln | 1 |

x xdx dxx x

x dxx x

dxx

x x C

3. 3sec ( )x dx

3 2

2

sec ( ) sec ( )sec( )

Integration by Partssec( ) sec ( )

sec( ) tan( ) tan( )

x dx x x dx

u x dv x dxdu x x v x dx

3 3

3

3 2

2

3

3

3

sec ( ) tan ( )sec( )

sec ( ) sec( ) tan( ) (sec ( ) 1)sec( )

sec( )sec ( ) sec ( )

sec ( )

tan( ) sec( )

sec( ) tan( ) ln | sec( ) tan( ) |

2

sec ( )

se

se

c( ) tan(

c

)

( sec( ))

x dx x dx

x dx

x dx x x dx

x dx x x x x dx

x x x dx

x x xx

x x

d x C

x dx x

x

3sec ( )

tan( ) ln | sec( ) tan( ) |

1 (sec( ) tan( ) ln | sec( ) tan( ) |)2

x

x x x C

x x x Cd xx

4. ln( )x dx

Integration by Partsln( )

ln( )

ln( )

1

ln( )

ln( )

u x dv dx

du

x

x

dx

dx v xx

x dx

x x C

x dx

x

Trig Substitution 

L. Marizza A. Bailey 

Example 1.  

2

14

dxx x+∫  

 Let x =  2) then 2 2sec (ta )n( dx dθ θ θ= . 

2

2

2

1 sec ( )2 tan( ) 4 4 tan ( )

sec ( )2 tan( )2sec( )

1 sec( ) cot( )41 1 cos( )4 cos( ) sin( )1 csc( )4

1 ln(csc( ) cot( )) + C4

d

d

d

d

d

θ θθ θ

θ θθ θ

θ θ θ

θ θθ θ

θ θ

θ θ

+

=

=

=

=

= − +

∫

∫

∫

∫

∫

 

    

Since x =2 tan(Θ), then tanΘ =  . 

Therefore  24 2csc( ) and cot( )x

xxθ θ+

= =  

Hence,  the integral is  

 2

2

1 ln(csc( ) cot( )) + C41 4 2= ln( )

41 1ln( 4 2) ln( ) C

4 4

x Cx

x x

θ θ= − +

− + ++

−= + + + +

 

     

Notes:  

• If you see  2 , you will need to use tan( )a x x a θ+ =  

• If you see  2 , you will need to use sin( )a x x a θ− =  

• If you see  2 , you will need to use sec( )x x aa θ=−  

• Remember the magic box:  tan( ) implies arctan( )y y xx ==  

• After you finish integrating with respect to theta, take your trig substitution and solve for the trig function. This will help you draw your triangle. 

     

Θ 

4

2 

x 

SOME MORE TECHNIQUES OF INTEGRATION

L. MARIZZA A. BAILEY

1. Partial Fractions

Suppose your model requires for population with limited resources:

The rate of increase of the population at time t is directly pro-portional to both P and L-P, where L is the maximum size of thepopulation

This is called the logistic growth model. This is modeled by the equation

dP

dt= kP (L− P ).

To find the general solution to the differential equation, we must separate thevariables:

dP

P (L− P )= kdt

Now we need to integrate. If we use u-substitution on the left side, we would use

u = P (L− P ) = PL− P 2,

then du = L− 2P . This, however, does not seem to help at all. This is why weneed to split up the fraction . It would be nice if there existed A,B ∈ R suchthat

1P (L− P )

=A

P+

B

L− P

Before we tackle the problem above, let us look at simpler rational functions.

Consider f(x) =1

(x− 1)(2x− 3).

How do we find,

f(x) =∫

1(x− 1)(2x− 3)

?

The same problem arises. We need to find A,B such that

1(x− 1)(2x− 3)

=A

x− 1+

B

2x− 3

By multiplying both sides by (x− 1)(2x− 3) we get

1 = A(2x− 3) + B(x− 1)

Date: February 02, 2008.

1

2

If this A and B exist, then they make the equation true regardless of x. Inparticular, this would be true for x = 1,

1 = A(2(1)− 3) + B(1− 1)1 = −A

Now we know that A = −1.Similarly by allowing x = 0, we get

1 = A(2(0)− 3) + B(0− 1)

1 = (−1)(−3) + B(−1)1 = 3−B

−2 = −B

2 = B

Therefore, we have

1(x− 1)(2x− 3)

=−1

x− 1+

22x− 3

To check your answer, plug our A and B into the equation

1 = A(2x− 3) + B(x− 1)

= (−1)(2x− 3) + 2(x− 1)= −2x + 3 + 2x− 2= 1

Now that we have split up the fraction we can integrate:

∫1

(x− 1)(2x− 3)dx =

∫ −1x− 1

dx +2

2x− 3dx

= − ln(x− 1) + ln(2x− 3)

This technique is called partial fractions.Try to solve the following integral using partial fractions:

Problem 1.

∫4

(x− 1)(x− 4)dx

3

Now let us see how to solve the logistic growth model.

1P (L− P )

=A

P+

B

L− P

1 = A(L− P ) + BP

Allowing P = L (remember L is just a number), and P = 0 would make ourcomputations much easier.If P = L,

1 = BL

or

1L

= B

If P = 0,

1 = A(L)

or

1L

= A

Hence,

1P (L− P )

= (1L

)(1P

+1

L− P)

Now to solve for P

∫1

P (L− P )dP =

∫kdt

1L

∫1P

dP +∫

1L− P

dP =∫

kdt

1L

ln(P )− ln(L− P ) = kt + C

Now we begin the grueling process of solving for P. Remember L is the limitingfactor, and k and C are constants.

ln(P

L− P) = Lkt + C

P

L− P= CeLkt

L− P

P=

C

eLkt

4

In the last step, we took the reciprocal of both sides in order to be able to breakthe fraction up into more pieces.

L

P− 1 =

C

eLkt

L

P=

C

eLkt+ 1

L

P=

C + eLkt

ekLt

P

L=

ekLt

C + ekLt

P =LekLt

C + ekLt

Now, we factor out ekLt from the numerator and denominator of the right handside,

P =L

Ce−kLt + 1

Given two initial conditions, we can solve for k and C.Let us attempt to apply this model to the following problem:

Problem 2. In a town of 100, 000, there were 20000 residents which heard of aradio announcement about the local political scandal. The rate of growth of thespread of information about the scandal is jointly proportional to the numberof people who heard it and the number of people who had not heard it. If 50%of the population heard about the scandal after one hour, how long was it until80% of the population heard it?

(a) Step 1: Write the differential equation.

(b) Step 2: Separate Variables.

(c) Step 3: Use Partial Fractions to write in an integrable form.

5

(d) Step 4: Integrate Both Sides. [ Don’t forget the C]

(e) Step 5: Solve for P .

(f) Step 6: Plug in your initial conditions.

6

2. Integration by Parts

The velocity of a particle is given by v(t) = (cos(t), tet). What is the positionfunction of the particle with respect to time if the initial position of the particlewas (1, 2)?

In order to solve this problem, we need to find the anti-derivative of both they and x velocity functions.

x(t) =∫

cos(t)dt

= sin(t) + C

Since at t = 0 the x coordinate was 1, we can substitute these values:

1 = sin(0) + C

implies C = 1. Therefore, the x-coordinate function is given by x(t) = sin(t)+C.Now, let us solve for the y-position function.

y(t) =∫

tetdt

Can we use u-substitution? Partial Fractions? This looks like the product oftwo functions. Can we use product rule? Let us review product rule, and see ifit gives us any ideas.

d

dx(uv) =

du

dxv +

dv

dxu

Integrating both sides, we achieve

uv =∫

vdu +∫

udv

Solving for one of the integrals,∫

udv = uv −∫

vdu

This formula is called Integration by Parts .How can we use this to help us solve the integral above?

Let u = t and dv = et.Then du = dt and v = et.Then by the formula,

∫udv =

∫tetdt

= tet −∫

etdt

= tet − et + C

By plugging in are initial condition, t = 0, then y = 2, we get

2 = y(0) = 0e0 − e0 + C

Simplification yields2 = C − 1

7

Thus C = 3 and y(t) = tet − et + 3. The position function is given by

(x(t), y(t)) = (sin(t) + 1, tet − et + 3)

Department of Mathematics, Arkansas School of Mathematics, Sciences and theArts

E-mail address: baileym@asmsa.com

TRIG TECHNIQUES

L. MARIZZA A. BAILEY

1. Trig Identities

Here are some examples that you may encounter while solving integrals withtrigonometric functions.

Problem 1. ∫sin3(x) cos2(x)dx

Solution 1. If sin3(x) were a sin(x), we could use u-substitution. Alas, it isnot. So, we will substitute 1 − cos2(x) for a sin2(x) and leave the sin(x) as anodd man out. We will be left with a polynomial in cos(x), and a sin(x)dx termperfectly positioned to be a du for u = cos(x).∫

sin3(x) cos2(x)dx =∫

sin(x)(1− cos2(x)) cos2(x)dx

=∫

sin(x)(cos2(x)− cos4(x))dx

= −∫

u2 − u4du letting u = cos(x) and du = − sin(x)dx.

= −u3

3+

u5

5+ C

= −cos3(x)3

+cos5(x)

5+ C

Problem 2. ∫sin2(x) cos2(x)dx

Solution 2. Unfortunately, writing sin2(x) = 1−cos2(x) will only leave us witha mess of cos2(x) to which we would have to individually decrease the exponentby using the half angle trig identity.∫

sin2(x) cos2(x)dx =∫

1− cos(2x)2

· 1 + cos(2x)2

dx

=14

∫1− cos2(2x)dx

=14

∫1− 1 + cos(4x)

2dx

=14

∫12− 1

2cos(4x)dx

=1x

8− sin(4x)

32+ C

Date: February 12, 2008.

1

2

Problem 3. ∫tan(x) sec4 xdx

Solution 3. This integral would be much friendlier if there was a tangentnearby, so that u = tan(x) and du = sec2(x)dx. If I break the integrand intosec2(x)(x) sec2(x) and substitute sec2(x) = (1 + tan2(x)), I will only have asec2(x) left. Let us proceed, and see what happens.∫

tan(x) sec4(x)dx =∫

tan(x)(1 + tan2(x)) sec2(x)dx

=∫

(tan(x) + tan3(x)) sec2(x)dx Let u = tan(x) then du = sec2(x)

=∫

u + u3du

=tan2(x)

2+

tan4(x)4

+ C

Problem 4. ∫sec3(x)dx

Solution 4. This integral is rather difficult in the sense that substituting asec2(x) = 1 + tan2(x) will not help because there is only one secant left, and itwould not help to make a u-substitution for u = tan(x). So instead we will tryintegration by parts. Let u = sec(x) and dv = sec( x)dx.Then du = sec(x) tan(x) and v = tan(x).

∫sec3(x)dx = sec(x) tan(x)−

∫tan2(x) sec(x)dx

Let u = tan(x) and dv = tan(x) sec(x)dx. Then du = sec(x) tan(x) and v =sec(x). ∫

sec3(x) = sec(x) tan(x)− [tan(x) sec(x)−∫

sec2(x) tan(x)dx]∫

sec3(x) = sec(x) tan(x)− tan(x) sec(x) +∫

sec(x) sec(x) tan(x)dx

We can use u-substitution.Let u = sec(x), then du = sec(x) tan(x)dx. Then∫

sec3(x) = sec(x) tan(x)− tan(x) sec(x) +∫

udu

∫sec3(x) = sec(x) tan(x)− tan(x) sec(x) +

sec2(x)2

+ C

The bottom line is that you want to use your brain and creativity to solvethese integrals, and most importantly, you need to know your trig identities.

Department of Mathematics, Arkansas School of Mathematics and Sciences.E-mail address: baileym@asmsa.org

Techniques of Integration and Differential Equation Worksheet

Name___________________________________

Evaluate the integral.

1) 2xex dx∫Integration by Parts:

2) x7 sec x8∫ dx

U - Substitution:

3)6 csc3 x

tan xdx∫

Trigonometric Identities:

4)1

0

dx

64 - x2∫

Trigonometric Substitution:

1

5)ln 3

0

et dt

16 + e2t∫

U - substitution, then trigonometric substitution: Donʹt forget to change the limits!!!!

Integrate the function.

6)1

-1

6

1 + 36t2 dt∫

Trigonometric Substitution: Remember to make a referenece triangle and rewrite the integratl in terms of t before

evaluating:

7)π/2

0

cos2 3x sin3 3x dx∫

Trigonometric Identities:

2

Express the integrand as a sum of partial fractions and evaluate the integral.

8)x + 9

x2 + 5x dx∫

Partial Fractions:

9)9

8

12

 x2 - 36 dx∫

Partial Fractions:

10)36 dx

x3 - 9x∫Partial Fractions

3

Answer KeyTestname: UNTITLED1

1) 2xex - 2ex + C

2)1

8 ln sec x8 + tan x8  + C

3) - 2 csc3 x + C

4) sin-1 1

8

5) 0.100

6) 2tan-1 6

7)2

45

8)1

5 ln 

x9

(x + 5)4 + C

9) 0.336

10) -4 ln x  + 2ln x - 3  + 2ln x + 3  + C

4