Download Week 2 Lab Name___________________ Part 1: Rules for

Week 2 Lab
Name___________________
Part 1: Rules for integrating trigonometric functions
 sin
m
x cos n x dx
There are three rules governing how we deal with integrals of this form.
1. If the power of the sine is odd and positive, save one sine factor and
convert the remaining factors to cosine. Then, expand and integrate.
2. If the power of cosine is odd and positive, save one cosine factor and
convert the rest to sine. Then, expand and integrate.
3. If the powers of both the sine and cosine are even and nonnegative, make repeated
use of the identities
sin 2 x 
1. Integrate
1  cos 2 x
1  cos 2 x
and cos 2 x 
2
2
. Which of the rules above will you choose? Why?
2. Choose a rule and rewrite the above integral using it. Be sure to put the factor you are saving to the
immediate left of the dx. Please underline the saved factor.
3. Convert the non-underlined terms in the result of part two into sin or cos depending on the rule you
chose to use. What identity can we use to do this?
4. What relationships between sin and cos are we taking advantage of when we use the rules above?
Based on step 2 above what should we set u equal to so that we can make a substitution that allows us
to use the general power rule to complete the integration?
u=
du =
5. Substitute u and du into your integral and use the general power rule to integrate.
6. Substitute the trig factor back in for u to write your final answer.
7. Use the trig rules to integrate
7. Use the trig rules to integrate
Integrals involving tan(x), sec(x), cot(x), csc(x)
1. If the power of sec or csc is even and positive, save a sec or csc squared
factor and convert the remaining factors to tangents.
2. If the power of the tan or cot is odd and positive, save a sectan or csccot
factor and convert the remaining factors to secants.
3. If there are no secant factors and the power of the tangent is even and
positive convert a tangent squared factor to secants, expand and repeat if
needed.
4. If the integral is of the form
 sec
m
x dx where m is odd and positive, use
integration by parts.
5. If none of the first four cases apply, try converting to sines and cosines.
1. Integrate
. Which of the rules above will you choose? Why?
2. Choose a rule and rewrite the above integral using it. Be sure to put the factor you are saving to
immediate left of the dx. Please underline the saved factor.
3. Convert the non-underlined terms in the result of part two into sec or tan depending on the rule you
chose to use. What identity can we use to do this?
4. What relationships between sec and tan are we taking advantage of when we use the rules above?
Based on step 2 above what should we set u equal to so that we can make a substitution that allows us
to use the general power rule to complete the integration?
u=
du =
5. Substitute u and du into your integral and use the general power rule to integrate. You may have to
use the distributive property to eliminate any parentheses first.
6. Substitute the trig factor back in for u to write your final answer.
7. Use the trig rules to integrate
7. Use the trig rules to integrate
Part II: Integration by Parts
Formula:
1. Integrate
. What should u be? What about dv? Why?
u=
dv =
2. Find du and v. Remember that du is the derivative of u and v is the Antiderivative of dv.
du =
v=
3. Substitute the values from steps 1 and 2 into the by parts formula. Can we integrate the new
integral
using basic formulas? Why or Why not? If not repeat steps 1 – 3 for the new integral.
=
4. In the above problem you will need to use integration by parts twice. You will have 3 terms in your
final answer. What is your final answer? Hint: Be sure to track through your negative signs.
5. In the above integral you had a finite number of times you had to use integration by parts. In the
integral
you could use integration by parts indefinitely. How can we avoid this?
6.
7.
Part III: Trigonometric Substitution
Trigonometric Substitution Rules
1. For
2. For
3. For
1. Integrate
let u =
let u =
let u =
and
and
and
=
=
=
. Identify the following:
a =5
u = 2x
du =2dx
What adjustment, if any, needs to be made to the integral to complete the substitution in to u?
The whole integral is to be divided by 2 as given in the next step.
2. Rewrite the above integral so that it is all in u.

dx
25  4 x
2

1
du

2
2
5  u2
3. Which of the trig substitution rules above will we use to solve this integral? Why? Draw and label the
relevant right triangle diagram. See the week 2 lecture or Section 28.8, page 858-9 for ideas on how to
do this.
We will use u=atanθ to solve this integral as it will help us in removing square root in denominator.
4. Use the rule you chose and the diagram to identify what you will replace u, du and
u  5tan 
du  5sec 2  d
a2  u 2  52  52 tan 2   5 1  tan 2   5sec
5. Rewrite the integral from step 2 using the replacements from step 4. Did you use all the
replacements? If not, why not? Can this integral be simplified? If so, do it?
1
du
1 5sec 2  d

2  52  u 2
2  5sec 

1
sec  d
2
6. Integrate the resulting trig function using a basic formula.
1
1
sec  d  ln(sec   tan  )  k

2
2
with.
7. Use the information from steps 1 and 4 to rewrite the answer in step 6 in terms of the variable x.
tan  
u
5
sec   1  tan 2   1 
u2
u 2  25

25
5
1
1  u 2  25 u 
1
1
Hence, (sec   tan  )  k  ln 
   k  ln  u 2  25  u   ln 5  k
 2
2
2 
5
5 
2 

1  2
ln u  25  u   k ' (Another constant)

2 
1
1
u  2 x. Hence, ln  u 2  25  u   k '  ln  4 x 2  25  4 x   k '


2 
2 
dx
1
 ln  4 x 2  25  4 x   k '
Hence, 

25  4 x 2 2 
8. How does knowing the trigonometric relationships as defined by the three sides of a right triangle
and the Pythagorean Theorem help us solve integrals using trig substitutions?
Now try these
9.
We substitute so that square root is removed from denominator.
x
dx
49  9 x 2

If we substitute
dx
x  7 1
9 2
x
49
9 2
x  sin 2  , we will get 1  sin 2   cos in denominator and we will be able to
49
remove square root from denominator. Hence, we substitute
9 2
49 2
7
x  sin 2   x 
sin   sin  .
49
9
3
Hence, dx 
7
cos  d
3
Hence, 

=
dx
x 49  9 x 2
7
cos 
3

7
49
sin  49  9  sin 2 
3
9
d
cos 
1
1
1
d  
d   cos ec d
7 sin  cos 
7 sin 
7
1
ln(cos ec  cot  )  k
7
as  cos ec d  ln(cos ec  cot  )
1  1
cos  
1 1  1  sin 2 
 ln 


k

ln 
7  sin  sin  
7 
sin 

9 x2
1  1 
1
49
 ln 
3
x
7 

7



k



1  7  49  9 x 2
 ln 
7 
3x
as sin  

k

3x
7

k

10.
We substitute so that square root is removed from denominator.
We substitute x 
We have dx 
Hence,
1
4

sec 
as sec2θ-1=tan2θ.
4
sec  tan 
d .
4
dx
16 x 2  1

= ln(sec   tan  )  k
sec  tan 
sec  tan 
d
d
sec  tan 
1
4
4


d   sec  d
4 tan 
4
sec2 
sec2   1
16 
1
16
as  sec  d  ln(sec   tan  )
1
4

1
4

= ln sec   sec 2   1   k

= ln  4 x  16 x 2  1   k as secθ=4x
