Download Integrals of trig functions

MATH 109 HANDOUT 1, INTEGRALS OF TRIGONOMETRIC FUNCTIONS
Evaluating
R
sinm x cosn dx.
1. If n is odd, we save one factor of cos x, substitute u = sin x, and express the integrand in terms of u:
Z
Z
sin x cos dx =
m
sinm x cosn−1 x cos x dx
n
Z
=
sinm x(1 − sin2 x)(n−1)/2 cos x dx
Z
=
um (1 − u2 )(n−1)/2 du
2. If m is odd, the same strategy as above works, save one factor of sin x and substitute u = cos x.
3. If both m and n are even, use double-angle formulae and go back to either case 1 or 2.
Evaluating
R
tanm x secn dx.
1. If n is even, save a factor of sec2 x, substitute u = tan x, and express the remaining factors in terms of
tan x:
Z
Z
tanm x secn dx =
tanm x secn−2 x sec2 x dx
Z
=
tanm x(1 + tan2 x)(n−2)/2 sec2 x dx
Z
=
um (1 + u2 )(n−2)/2 du
2. If m is odd, save a factor of sec x tan x, substitute u = sec x, and express the remaining factors in
terms of sec x:
Z
Z
tanm x secn dx =
tanm−1 x secn−1 x tan x sec x dx
Z
=
(sec2 x − 1)(m−1)/2 secn−1 xtan x sec x dx
Z
=
Evaluating
R
(u2 − 1)(m−1)/2 un−1 du
sec x dx.
This shows in the textbook on top of page 483, following the second approach I gave in the lecture. Let me
1
2
MATH 109 HANDOUT 1, INTEGRALS OF TRIGONOMETRIC FUNCTIONS
finish the first approach here:
Z
Z
sec x dx =
Z
=
Z
=
Z
=
Z
=
1
dx
cos x
cos x
dx
cos2 x
cos x dx
(let u = sin x, then du = cos x dx)
1 − sin2 x
du
(partial fraction trick, we will learn it)
1 − u2
1
1
1
du
+
2 1−u 1+u
Z
Z
1
1
1
1
=
du +
du (how to do these two integrals?)
2
1−u
2
1+u
1
1
= − ln|1 − u| + ln|1 + u| + C
2
2
1 1+u
|+C
= ln|
2 1−u
r
1+u
= ln
+C
1−u
r
1 + sin x
= ln
+ C (check the final step)
1 − sin x
= ln|sec x + tan x| + C