For Practice:
a.
b.
c.
d.
π
3 cos5x sin x dx
0
∫
∫3x e dx
2
e.
x
f.
g.
Sec 7.2: Trigonometric Integrals
This section involves integrals that are combinations of trig functions.
To evaluate them, you will need to look for patterns in the powers and
use trig identities to make substitutions.
Some identities you will be using (see these on p. 2 of the reference pages in the front of your text.)
sin2x + cos2x = 1
sin 2x = 2 sin x cos x
1 + tan2x = sec2x
2
cos 2x = cos2x ­ sin2x = 2cos2x ­ 1 = 1 ­ 2sin x 1 + cot2x = csc2x
sin2x = cos2x = We'll start with integrals involving sines and cosines.
EXAMPLE: Evaluate each of the following integrals.
(a)
∫sin2x cosx dx
(b)
∫sin2x cos3x dx
Strategy for evaluating ∫sinmx cosnx dx:
1. If one of the powers is odd, save one factor (cos x or
sin x) and substitute for an even power using 1- cos2x or 1sin2x so that the rest of the integrand is all the other
function.
2. If you have only even powers, substitute using the
double angle formulas for cos2x or sin2x.
EXAMPLE: Evaluate each of the following integrals.
(a)
∫sin2x cos5x dx
(b)
∫sin3x cos4x dx
EXAMPLE: Evaluate ∫cos2x dx
What about integrals that have trig functions other than sine and
cosine? Often similar strategies can be followed - (see p.474 for more
details on how to handle secants and tangents.)
a. If power of sec is even, save a factor of sec2x and use
sec2x = 1 + tan2x to express the remaining factors in terms of tan x.
Then substitute u = tan x.
b. If power if tan is odd, save a factor of secxtanx and use
tan2x = sec2x - 1 to express the remaining factors in terms of sec x.
Then substitute u = sec x.
Recall:
∫ tanxdx = ln|secx| + C
∫secxdx = ln|secx + tanx| + C
EXAMPLE: Evaluate ∫tan x sec4x dx
and
In each of the examples thus far, the arguments have been
the same for all the trig functions - that is, we had tanx,
secx, cosx, etc. What if the arguments differ, as in the
following example? Another trig identity will be helpful
here.
See p.476 for details.
EXAMPLE: Evaluate ∫sin6x sin5x dx