Download Lecture Notes for Section 6.3

Calc 2 Lecture Notes
Section 6.3
Page 1 of 8
Section 6.3: Trigonometric Techniques of Integration
Big idea: A lot of wicked-looking integrals can be computed using trigonometric identities and
substitutions.
Big skill:. You should be able to find the antiderivative of products of powers of sine and cosine
functions and tangent and secant functions, as well as make an appropriate substitution for
integrands of the form
a2  x2 and
x2  a2 .
I. Integrals of the form  sin m  x  cos n  x  dx .
1. For integrals of this form, you will either integrate by substitution, letting
u  x   sin  x  or u  x   cos  x  , or you will use a reduction formula. Which path
you take depends on whether m or n is even or odd…
m
n
u(x)
Odd
Even
u  x   cos  x  ; see Case 1
Even
Odd
Odd
Even
Odd
Even
u  x   sin  x  ; see Case 1
Either one; see Case 1
Use reduction formulas; see Case 2
2. Case 1 of  sin m  x  cos n  x  dx : m or n an odd positive integer.
i. Isolate one factor of the odd power to serve as the differential for
integration by substitution.
ii. Replace the remaining even powers using the Pythagorean identity:
sin 2   cos 2   1
iii. Integrate by substitution.
iv. Practice:  sin 4  x  cos3  x  dx 
v. Practice:  sin 3  x  cos 2  x  dx 
Calc 2 Lecture Notes
Section 6.3
Page 2 of 8
vi. Practice:  sin 3  x  cos3  x  dx 
3. Case 2 of  sin m  x  cos n  x  dx : m and n both even positive integers.
i. Use the power reduction identities:
1
1
sin 2  x   1  cos  2 x   or cos 2  x   1  cos  2 x   .
2
2
2
ii. Practice:  sin  x  dx 
iii. Practice:  sin 4  x  dx 
Calc 2 Lecture Notes
Section 6.3
Page 3 of 8
iv. Practice:  sin 2  x  cos 2  x  dx 
4. Don’t forget: You can also use the power reduction integral formulas:
1
n 1
cos n  2  x  dx
i.  cos n  x  dx  cos n 1  x  sin  x  
n
n 
1
n 1
sin n  2  x  dx
ii.  sin n  x  dx   sin n 1  x  cos  x  

n
n
4
iii. Practice:  sin  x  dx 
Calc 2 Lecture Notes
II. Integrals of the form
Section 6.3
Page 4 of 8
 tan  x  sec  x  dx .
m
n
1. For integrals of this form, you will either integrate by substitution, letting
u  x   tan  x  or u  x   sec  x  , or you will use a reduction formula. Which path
you take depends on whether m or n is even or odd…
m
n
u(x)
Odd
Even
u  x   sec  x  ; see Case 1
Or u  x   tan  x  ; see Case 2
Odd
Odd
Even
Even
Odd
Even
u  x   sec  x  ; see Case 1
Use reduction formulas; see Case 3
u  x   tan  x  ; see Case 2
2. These cases revolve around the basic integrals  sec2  x  dx  tan  x   c and
3.
 sec  x  tan  x  dx  sec  x   c .
Case 1 of  tan  x  sec  x  dx : m is an odd positive integer.
m
n
i. Isolate one factor of sec(x)tan(x) for the differential.
ii. Replace the remaining even powers of tan(x) using the Pythagorean
identity: tan 2 x  sec 2 x  1 .
iii. Integrate by substitution ( u(x) = sec(x) ).
iv. Practice:  tan 3  x  sec2  x  dx 
Calc 2 Lecture Notes
4. Case 2 of
Section 6.3
Page 5 of 8
 tan  x  sec  x  dx : n is an even positive integer.
m
n
i. Isolate one factor of sec2(x) for the differential.
ii. Replace the remaining even powers of sec(x) using the Pythagorean
identity: sec 2 x  1  tan 2 x
iii. Integrate by substitution ( u(x) = tan(x) ).
iv. Practice:  tan 4  x  sec4  x  dx 
5. Case 3 of
 tan  x  sec  x  dx : m is an even positive integer and n is an odd
m
n
positive integer.
i. Replace the even powers of tan(x) using the Pythagorean identity:
tan 2 x  sec 2 x  1 .
ii. Use the following reduction formula to integrate the powers of sec(x):
1
n2
n
n2
n2
 sec  x  dx  n  1 sec  x  tan  x   n  1  sec  x  dx
(This can be proven using integration by parts where u  x   secn2  x 
and dv  sec2  x  dx )
iii. Since n is odd, you will always be left with having to integrate sec(x),
which can be done using the following formula:
tan  x   sec  x 
 sec  x  dx   sec  x   tan  x   sec  x  dx
sec  x  tan  x   sec 2  x 

dx
tan  x   sec  x 

1
d
 sec  x   tan  x   dx
tan  x   sec  x  dx
 ln sec  x   tan  x   c
iv. Practice:
 tan  x  sec  x  dx 
2
3
Calc 2 Lecture Notes
Section 6.3
III. Integrals involving integrands with
a2  x2 or
Page 6 of 8
x2  a2 .
1. Case 1: Integrals involving a2  x2
i. Check to see if you can use substitution or integration by parts.
ii. If not, make the substitution x  a sin   , and dx  a cos   d .
iii. Simplify the radical using algebra and the Pythagorean identity:
a 2   a sin     a 2 1  sin 2     a 2 cos 2    a cos  
2
iv. Integrate.
v. To back-substitute, use the following trick from trigonometry.
x
Since x  a sin   , or sin    , we can draw the diagram below to
a
evaluate any trigonometric function of :
vi. Practice:
x
1
5  x2
dx 
Calc 2 Lecture Notes
Section 6.3
Page 7 of 8
2. Case 2: Integrals involving a2  x2
i. Check to see if you can use substitution or integration by parts.
ii. If not, make the substitution x  a tan   and dx  a sec2   d .
iii. Simplify the radical using algebra and the Pythagorean identity:
a 2   a tan     a 2 1  tan 2     a 2 sec2    a sec  
2
iv. Integrate.
v. To back-substitute, use the following trick from trigonometry.
x
Since x  a tan   , or tan    , we can draw the diagram below to
a
evaluate any trigonometric function of :
vi. Practice:
x
1
5  x2
dx 
Calc 2 Lecture Notes
Section 6.3
Page 8 of 8
3. Case 3: Integrals involving x2  a2
i. Check to see if you can use substitution or integration by parts.
ii. If not, make the substitution x  a sec   and dx  a sec   tan   d .
iii. Simplify the radical using algebra and the Pythagorean identity:
 a sec   
2
 a 2  a 2  sec2    1  a 2 tan 2    a tan  
iv. Integrate.
v. To back-substitute, use the following trick from trigonometry.
x
Since x  a sec   or sec    , we can draw the diagram below to
a
evaluate any trigonometric function of :
vi. Practice:
x
1
x2  5
dx 