Download Section 6.2

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Section 6.2: Trigonometric Integrals
Practice HW from Larson Textbook (not to hand in)
p. 382 # 5-13 odd, 19-31 odd, 51-55 odd
Integrals Involving Powers of Sine and Cosine
Example 1: Integrate  sin 5 x cos x dx
Solution:
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Larger Powers of Sine and Cosine
Two Types
1. Odd Powers of Sine and Cosine both larger than one: Attempt to write the sine or
cosine term with the lowest odd power in terms of an odd power times the square of
sine and cosine. Then rewrite the squared term using the Pythagorean identity
sin 2 u  cos 2 u  1 .
2. Only Even Powers of Sine and Cosine: Use the identities
sin 2 u 
1  cos 2u 1
 (1  cos 2u )
2
2
and
cos 2 u 
Note in these formulas the initial angle is always doubled.
1  cos 2u 1
 (1  cos 2u ) .
2
2
2
Example 2: Integrate  cos 5 x sin 3 x dx
Solution:
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Example 3: Integrate  cos 3 x sin 2 x dx
Solution: Here, the lowest odd power trigonometric term is cos 3 x . Starting with this
fact, we integrate the function as follows:
 cos
3
x sin 2 x dx   cos 2 x cos x sin 2 x dx
(Use fact that cos 3 x  cos 2 x  cos x )
  (1  sin 2 x) cos x sin 2 x dx
(Substitut e cos 2 x  1  sin 2 x )
  (cos x sin 2 x  cos x sin 4 x) dx
(Distribut e cos x sin 2 x to (1  sin 2 x))
  (sin 2 x cos x  sin 4 x cos x) dx
(Rearrange terms)
  sin 2 x cos x dx   sin 4 x cos x dx
(Break into separate integrals)
(Use u  du substituti on on each integral)
Let u  sin x, du  cos x dx
  u 2 du   u 4 du
(Make substituti on0
1
1
 u3  u5  C
3
5
(Integrate )
1
1
 (sin x) 3  (sin x) 5  C
3
5
(Substitut e for u )
1
1
 sin 3 x  sin 5 x  C
3
5
(Rewrite in standard trig form)
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Example 4: Integrate  sin 3 2 x dx
Solution:
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5

2
Example 5: Integrate
 sin
2
3 x dx
0
Solution: Since the sine term is even, we use the identities sin 2 u 
1
(1  cos 2u ) to
2
rewrite the integral. Hence we have


2
2
2
 sin 3x dx 

0
0
1
(1  cos 6 x) dx
2
Use fact that sin 2 3 x 
1
(1  cos 6 x)
2

2
1
1
1
(Distribut e the )
2
 ( 2  2 cos 6 x) dx

0


2


0
2
1
1
dx   cos 6 x dx
2
2
0
(Break into separate integrals)

1 1
1
2
  x  ( ) sin 6 x 
2 6
2
0
(By u  du substituti on,  cos 6 x dx 
1
sin 6 x  C )
6

1
1
2
  x  sin 6 x 
12
2
0
1
  1
1
1 

  ( )  sin 6( )    (0)  sin 6(0) 
2  2
12
 2 2 12

1
 1
 

   sin 3    0  sin 0 
12
 4 12
 

1
 1
 

   (0)    0  (0) 
12 
 4 12  



4
(Note that sin 3  0 and sin 0  0)
00

4
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Integrals Involving Secant and Tangent
Example 6: Integrate
 tan
3
x sec 2 x dx
Solution:
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Larger Powers of Tangent and Secant
Steps to Consider (listed in detail in textbook p. 379)
1. If the power of secant is even (2, 4, 6, 8, etc.), save sec 2 x and convert the remaining
factors to tangents.
2. If the power of tangent is odd and positive, save sec x tan x and convert the
remaining factors to secants.
3. If there are no secant factors and power of tangent is even (2, 4, 6, 8, etc.), convert
tan 2 u factor to sec 2 u factor and expand.
4. If integral is of the form  sec m x dx , where m is odd (3, 5, 7, 9, etc.), use integration
by parts (see Example 5 on p. 372).
5. If nothing works, convert to sines and cosines.
Useful Facts
1. sec 2 x  1  tan 2 x
d
tan x  sec 2 x
2.
dx
d
sec x  sec x tan x
3.
dx
4.  sec 2 x dx  tan x  C
5.
 sec x dx  ln | sec x  tan x | C
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Example 7: Integrate  sec 4 2 x dx
Solution:
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Example 8: Integrate
 tan
3
t sec 3 t dt
Solution: Note that there are no even powers of secant, so we skip step 1 of the listed
steps and proceed to step 2. Here, we save a sec t tan t and convert the rest of the terms
of secants. We proceed using the following steps:
 tan
t sec 3 t dt   sec t tan t sec 2 t tan 2 t dt
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  sec t tan t sec 2 t (sec 2 t  1) dt
(Substitut e tan 2 t  sec 2 t  1)
  sec t tan t sec 2 t  sec 2 t  sec t tan t sec 2 t  1) dt
  sec 4 t sec t tan t  sec 2 t sec t tan t ) dt
(Distribut e)
(Simplify)
  sec 4 t sec t tan t dt   sec 2 t sec t tan t ) dt
(Break into separate integrals)
Use u  du Substituti on
  u du   u du
4

2
1 5 1 3
u  u C
5
3
1
1
 sec 5 t  sec 3 t  C
5
3
Let u  sec t , du  sec t tan t dt
(Integrate )
(Write in terms of original variable)
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