Calculus II
Spring 2008
Recitation Session #6
(§8.4 and §8.5)
 Summary of important concepts
§8.4
[1] Products of powers of sines and cosines
 sin
m
x cosn xdx
Case studies:
(a) If the power over sine is odd, save a factor of sin x , use the trig-identity sin 2 x  1  cos2 x to rewrite sine
in terms of cosine, then let u  cos x . That factor sin x we saved earlier will be cancelled after usubstitution. When evaluating the u-integral, power rule for integration applies.
(b) If the power over cosine is odd, save a factor of cos x , use the trig-identity cos2 x  1  sin 2 x to rewrite
cosine in terms of sine, then let u  sin x . That factor cos x we saved earlier will be cancelled after usubstitution. When evaluating the u-integral, power rule for integration applies.
(c) When the power over sine and cosine are all even, be prepared for some algebra. Use the powerreducing/double angle identities to reduce the powers over sine and cosine to one.
Power-reducing/double-angle identities: cos2 A 
1  cos 2 A
,
2
sin 2 A 
1  cos 2 A
. Notice that the angle is
2
doubled on the right side of the identity, and the power is reduced from 2 to 1.
Sometimes, one has to use product to sum trig-identities to obtain a number of single trigonometric
functions, i.e. no product, and powers are all one over sines and cosines.
Product to sum trig-identities: cos A cos B 
sin A cos B 
1
1
cos( A  B)  cos( A  B) , sin Asin B   cos( A  B)  cos( A  B)
2
2
1
sin( A  B)  sin( A  B) .
2
Now, let’s take a look at an example:
  1  cos 4 x  1  cos 2 x 
1
(2 x) cos 2 xdx   

 dx   (1  cos 2 x  cos 4 x  cos 4 x cos 2 x)dx
2
2
4



1
1
1
1
1
1
1
1
  dx   cos 2 xdx   cos 4 xdx   cos 4 x cos 2 xdx  x  sin 2 x  sin 4 x    cos 6 x  cos 2 x  dx
4
4
4
4
4
8
16
8
1
1
1
1
1
1
1
1
1
1
 x  sin 2 x  sin 4 x   cos 6 xdx   cos 2 xdx  x  sin 2 x  sin 4 x  sin 6 x  sin 2 x  C
4
8
16
8
8
4
8
16
48
16
1
1
1
1
 x  sin 2 x  sin 4 x  sin 6 x  C
4
16
16
48
 sin
2
Your turn:
 sin
2
x cos4 xdx
[2] Eliminating the square root.
We already covered this in section 8.1. See example like the following in 8.1.

1  cos 4 xdx   2sin 2 2 xdx  2  sin 2 xdx  
2
cos 2 x  C
2
The main idea is to use half-angle identities to come up with a perfect square under the square root. The
half-angle identities are closely related to the double angle-identities:
Half-angle identities: 1  cos A  2cos2
A
A
, 1  cos A  2sin 2 . Notice that the angle on the right side is
2
2
halved. Do you see the connection between the half-angle identities and the double-angle identities?
Trigonometry is an easy subject if you see the connections among the trig-identities.
[3] The power over tangents and secants
Case studies:
(a) If the power over secant is even, save a factor of sec2 x , use the trig-identity sec2 x  1  tan 2 x to rewrite
sec x in terms of tan x , then let u  tan x . That factor sec2 x we saved earlier will be cancelled after usubstitution. When evaluating the u-integral, power rule for integration applies.
(b) If the power over tan x is odd, save a factor of sec x tan x , use the trig-identity tan 2 x  sec2 x  1 to rewrite
tan x in terms of sec x , then let u  sec x . That factor sec x tan x we saved earlier will be cancelled after usubstitution. When evaluating the u-integral, power rule for integration applies.
 tan
Example:
3
u sec x 1
1
 1

tan 2 x(sec x tan x)dx  
(sec2 x  1)(sec x tan x)dx   (u 2  1)du
 sec x
u
 sec x

xdx  

1
u2
sec2 x
  (u  )du 
 ln u  C 
 ln sec x  C
u
2
2

Your turn:
 sec
4
x tan 2 xdx
§8.5
Trigonometric substitutions require the following knowledge
(I) trig-identities



(II) the right-triangle scheme (simplify expressions such as sin  tan 1

 done in section 7.7)

x2  1 
2
(III) trigonometric integrals from section 8.4
Note: Update the integral limits when the integral is a definite integral. That way you do not do backsubstitution, as such part (b) above can be skipped.
Warning: you cannot do both: update the integral limits and do back-substitution. That will guarantee a
wrong answer.
Case studies:
(a) Whenever you see the expression a2  x2 appear in the integrand, use the following trig-substitution
x  a tan  , dx  a sec2  d
Make sure that  is the only variable seen in the new integrand after the substitution. You will use the trigidentity 1  tan2   sec2  and possibly other trig-identities to simplify the expressions.
(b) Whenever you see the expression a2  x2 appear in the integrand, use the following trig-substitution
x  a sin  , dx  a cos d
Make sure that  is the only variable seen in the new integrand after the substitution. You will use the trigidentity 1  sin2   cos2  and possibly other trig-identities to simplify the expressions.
(c) Whenever you see the expression x2  a 2 appear in the integrand, use the following trig-substitution
x  a sec , dx  a sec tan  d
Make sure that  is the only variable seen in the new integrand after the substitution. You will use the trigidentity sec2   1  tan 2  and possibly other trig-identities to simplify the expressions.
Note: if the coefficient of x 2 is not 1 or  1 , remember to factor out that coefficient so that the coefficient of
x 2 becomes 1 or  1 .
Next, we will use a definite integral to demonstrate trig-substitutions. Notice that the integral limits will be
updated, and there is no back-substitution.
Example:
4
4
x  2sec  1   / 3


dx
dx
1 
dx
2sec tan  d







 3
 3

 3
2
2
2
2 2 x x  4
2 0 8sec3  4sec 2   4
2 x 2 x  8 2 x 2( x  4)
4




1 
 /3

4 2
0
tan  d
sec2  4sec 2   4
 /3
1   / 3 d
1 



2
8 2 0 sec  8 2 0
1
 |0 / 3   1
 32 2
16 2 
 /3
0

1 
 /3

4 2
0
cos 2  d 
tan  d
sec 2  4(sec 2   1)

 /3

4 2
0
tan  d
sec 2  4 tan 2 
 /3
1   / 3 1  cos 2
1 
d 


2
8 2 0
16 2 0
cos 2 d (2 ) 

48 2

d 
1
16
 /3

2 0
cos 2 d


sin 2 |0 / 3     1  3  0     3


32 2
48 2 32 2  2
 48 2 64 2
1
2
6

96 128
 Homework questions:
§8.4: 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 25, 29, 33, 35, 37, 40, 41.
§8.5: 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 27.
.
1 
Calculus II
Recitation Session #6
(§8.4 and §8.5)
NAME: _______________________________
(20-points total)
(1-8 pts) Evaluate the following integrals:
(a-2pts)



1
dx
x 4
2
(b-6pts)  4sin 2 (2 x)cos2 x dx
(2-12pts) Evaluate the following integral by trigonometric substitution




x3 dx
2 x2  8