# Calculus

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```					                                                          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 u-
substitution. 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 u-
substitution. 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 power-
reducing/double angle identities to reduce the powers over sine and cosine to one.

1  cos 2 A                   1  cos 2 A
Power-reducing/double-angle identities: cos2 A                           ,     sin 2 A                . Notice that the angle is
2                             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.
1                                             1
Product to sum trig-identities: cos A cos B         cos( A  B)  cos( A  B) , sin Asin B   cos( A  B)  cos( A  B)
2                                             2
1
sin A cos B      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
 sin
(2 x)cos 2 xdx                                dx   (1  cos 2 x  cos 4 x  cos 4 x cos 2 x)dx
2

       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
x cos4 xdx
[2] Eliminating the square root.

We already covered this in section 8.1. See example like the following in 8.1.
2
   1  cos 4 xdx   2sin 2 2 xdx  2  sin 2 xdx  
2
cos 2 x  C

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:
A                     A
Half-angle identities: 1  cos A  2cos2                 , 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 sec 2 x we saved earlier will be cancelled after u-
substitution. 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 u-
substitution. When evaluating the u-integral, power rule for integration applies.

                                                                   u sec x 1
1                             1                                        
 tan       xdx            tan 2 x(sec x tan x)dx         (sec2 x  1)(sec x tan x)dx   (u 2  1)du
3
 sec x                           sec x                                    u
Example:
     1      u2              sec2 x
  (u  )du      ln u  C          ln sec x  C
     u       2                2

 sec
4
Your turn:                       x tan 2 xdx

§8.5

Trigonometric substitutions require the following knowledge

(I) trig-identities
             2     
(II) the right-triangle scheme (simplify expressions such as sin  tan 1                             done in section 7.7)
                   
            x2  1 
(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 back-
substitution, 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

Case studies:

(a) Whenever you see the expression a 2  x 2 appear in the integrand, use the following trig-substitution
x  a tan  , dx  a sec 2  d
Make sure that  is the only variable seen in the new integrand after the substitution. You will use the trig-
identity 1  tan 2   sec2  and possibly other trig-identities to simplify the expressions.

(b) Whenever you see the expression a 2  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 trig-
identity 1  sin 2   cos 2  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 trig-
identity 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:
x  2sec  1   / 3
4                  4             4
    dx           dx         1      dx                         2sec tan  d

 3                           
 3                     

 3                                        2 0 8sec3  4sec2   4
2 x 2 x  8 2 x 2( x  4)
2             2        2 2 x x  4
2

 /3                                       /3                                      /3
1               tan  d        1                          tan  d        1                     tan  d
                                                                              
4 2 0


sec2  4sec2   4 4 2 0

sec2  4(sec2   1) 4 2 0              sec2  4 tan 2 
 /3                                                         /3
1   / 3 d   1                                  1   / 3 1  cos 2        1                       1      /3
                                    cos 2  d                         d                   d           0      cos 2 d
8 2 0 sec  8 2 0
2
8 2 0          2          16 2 0                  16 2
1   /3   1             /3                                  1                           1          3           3
         |              0         cos 2 d (2 )                     sin 2 | / 3                       0    
 0  32 2                                                               0 
 2     48 2 64 2
16 2                                                 48 2       32 2                     48 2 32 2              
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.
.
Calculus II
Recitation Session #6
(§8.4 and §8.5)

NAME: _______________________________                      (20-points total)

(1-8 pts) Evaluate the following integrals:
     1
(a-2pts)           dx   (b-6pts)  4sin 2 (2 x)cos2 x dx
   x 4
2
(2-12pts) Evaluate the following integral by trigonometric substitution
   x3dx


   2 x2  8

```
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