Math 152, Fall 2008
Lecture 7.
09/16/2008
HW#3 is due Wednesday, September 17, 11:55 PM.
Example 1. Evaluate the integral.
(a)
π/4
R
sin4 x cos3 x dx
(b)
π/2
R
sin2 x cos2 x dx
(d)
0
(c)
0
(e)
R
cos4 x dx
R
(2x + 3)e x dx
π/3
R
0
x sin x cos x dx
Chapter 8. Techniques of integration
Section 8.2 Trigonometric integrals
How to evaluate
R
tanm x secn x dx
(a) if the power of secant is even n = 2k, save a factor of sec2 x
and use sec2 x = 1 + tan2 x to express the remaining factors in
terms of tan x:
Z
Z
m
2k
tan x sec x dx = tanm x(sec2 x)k−1 sec2 x dx =
Z
tanm x(1 + tan2 x)k−1 sec2 x dx
Then substitute u = tan x
Example 2. (a)
(c)
π/4
R
0
R
tan2 x dx
tan4 x sec2 x dx
(d)
(b)
R
R
sec4 x dx
tan3 x dx
(b) if the power of tangent is odd (m = 2s + 1), save a factor of
tan x sec x and use tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Z
Z
2s+1
n
tan
x sec x dx = (tan2 x)s secm x tan x sec x dx =
Z
(sec2 x − 1)s secm x tan x sec x dx
Then substitute u = sec x
Example 3.
R
(a) tan3 x sec3 x dx
(b)
R
sec x dx
(c)
R
csc3 x dx
To evaluate the integrals
(a)
Z
sin mx cos nx dx
(b)
Z
sin mx sin nx dx
(c)
Z
cos mx cos nx dx
use the corresponding identity:
1
[sin(α − β) + sin(α + β)]
2
1
(b) sin α sin β = [cos(α − β) − cos(α + β)]
2
1
(c) cos α cos β = [cos(α − β) + cos(α + β)]
2
Example 4.
R
R
(a) sin 5x sin 2x dx
(b) sin 3x cos x dx
R
(c) cos 3x cos 4x dx
(a) sin α cos β =
Section 8.3 Trigonometric substitution
We can make a substitution of the form x = g (t) by using the
Substitution Rule in reverse. Assume that g is one-to-one function
(g −1 exists). Then
Z
Z
f (x)dx = f (g (t))g ′ (t)dt
This kind of substitution is called inverse substitution.
Table of trigonometric substitutions
Expression
√
a2 − x 2
√
a2 + x 2
√
x 2 − a2
Substitution
x = a sin t, −π/2 ≤ t ≤ π/2
Identity
1 − sin2 t = cos2 t
x = a tan t, −π/2 < t < π/2
1 + tan2 t = sec2 t
x = a sec t, 0 ≤ t ≤ π/2 or π ≤ t ≤ 3π/2 sec2 t − 1 = tan2 t
Example 5.
Z
Z
dx
x3
√
√
dx
(a)
(b)
x2 + 4
x2 1 − x2
Z
dx
√
(d)
2
x + 4x + 8
(c)
Z
x2
dx
√
16x 2 − 9
Example 6. Use trigonometric substitution to show that
Z
p
dx
√
= ln(x + x 2 + a2 ) + C
x 2 + a2