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Jim Lamb ers Math 2B Fall Quarter 2004- Le ture 20 Notes
These notes orresp ond to Se tion 8.2 in the text.
Trigonometri Integrals
So far we have obtained several integration rules by reversing various di erentiation rules. Addi- tional integration rules an b e obtained by employing trigonometri identities to onvert integrands involving trigonometri fun tions into a form in whi h other integration rules an b e applied. For su h integrands, the resulting rules an then b e used dire tly, instead of ontinuing to rely on the trigonometri identities used to establish them. For example, supp ose we wish to evaluate an integral of the form Z sinm^ x os n^ x dx: (1)
If n is o dd, then we an use the identity sin^2 x + os 2 x = 1 to express n 1 of the p owers of osine in terms of sine. Then, we an use the substitution u = sin x. Be ause du = os x dx, the resulting integral with resp e t to u will b e simple to evaluate, sin e the integrand will onsist of several terms of the form um+2k^ , for k = 0 ; : : : ; (n 1)=2, whi h an easily b e integrated using the p ower rule. An integral of the form (1) where m is o dd an b e handled similarly, using the same identity.
Example 1 Evaluate (^) Z
sin^3 x os^2 x dx: (2)
Solution Sin e the p ower of sine is o dd, we write all but one p ower of sine in terms of osines: Z sin^3 x os 2 x dx =
Z
sin x(1 os^2 x) os 2 x dx =
Z
sin x os 2 x dx
Z
sin x os 4 x dx: (3)
Using the substitution u = os x, with du = sin x dx, yields Z sin x os 2 x dx
Z
sin x os 4 x dx =
Z
u^2 du +
Z
u^4 du =
u^5 5 ^
u^3 3 +^ C^ =^
os^5 x 5 ^
os 3 x 3 +^ C^ :^ (4)
2
If b oth m and n are even, then the half-angle formulas
sin^2 x = 1 2
(1 os 2 x); os^2 x = 1 2
(1 + os 2 x) (5)
an b e used to redu e the p owers of sine and osine until every term in the transformed integrand either has an o dd p ower of sine or osine, or until a simpler integration rule an b e used. If a term of the form sin x os x arises, this an b e handled using the substitution u = os x or u = sin x, but one may nd it easier to use the double-angle formula
sin x os x = 12 sin 2 x: (6)
Example 2 Evaluate (^) Z
sin^2 x os^2 x dx: (7)
Solution Using the double-angle formula
sin 2 x = 2 sin x os x (8)
yields (^) Z
sin^2 x os^2 x dx =
Z ^ sin 2 x 2
dx =
Z
sin^2 2 x dx: (9)
We an then use a half-angle formula
sin^2 x = 1 ^ os^2 x 2
to obtain Z sin^2 x os 2 x dx =
Z (^1) os 4 x 2 dx =
Z
1 os 4 x dx
= (^18)
x sin 4 4 x
= x 8 sin 32 4 x+ C : (11)
2
Integrals of the form (^) Z tan m^ x se n^ x dx (12)
Using the substitution u = se x, with du = se x tan x dx, yields Z se 4 x tan^3 x dx =
Z
se 6 x tan x dx
Z
se 4 x tan x dx
=
Z
se 5 x se x tan x dx
Z
se 3 x se x tan x dx
=
Z
u^5 du
Z
u^3 du
= u
u
+ C
= se^
(^6) x 6 ^
se 4 x 4 +^ C^ :^ (19)
2
Other ases are not as straightforward. It may b e ne essary to use other trigonometri identities, integration by parts, or the rules for integrating tan x and se x: Z tan x dx = ln j se xj + C ; (20) Z se x dx = ln j se x + tan xj + C : (21)
Example 4 Evaluate (^) Z
se 3 x tan 2 x dx: (22)
Solution Sin e the p ower of tangent is even and the p ower of se ant is o dd, it is not p ossible to rewrite p owers of one in terms of p owers of the other and use a simple substitution as in the previous example. Instead, we an pro eed using integration by parts with
u = se x tan 2 x; dv = se 2 x; (23)
whi h implies du = (2 se 3 x tan x + se x tan^3 x) dx; v = tan x: (24)
Before applying integration by parts, we use the identity tan^2 x = se 2 x 1 to write
du = (2 se 3 x tan x + se x(se 2 x 1) tan x) dx = (2 se 3 x tan x + se 2 x se x tan x se x tan x) dx = (3 se 3 x tan x se x tan x) dx: (25)
Integrating by parts, we obtain Z se 3 x tan^2 x dx = se x tan 3 x 3
Z
se 3 x tan 2 x dx +
Z
se x tan^2 x dx: (26)
Rearranging algebrai ally yields
4
Z
se 3 x tan^2 x dx = se x tan 3 x +
Z
se x tan 2 x dx (27)
or (^) Z se 3 x tan^2 x dx = 1 4
se x tan^3 x + 1 4
Z
se x tan^2 x dx: (28)
We now fo us on the remaining integral Z se x tan^2 x dx: (29)
Rewriting the p owers of tangent as p owers of se ant yields Z se x tan^2 x dx =
Z
se x(se 2 x 1) dx =
Z
se 3 x se x dx: (30)
To integrate se x, we pro eed as follows, using the substitution u = se x + tan x with du = (se x tan x + se 2 x) dx: Z se x dx =
Z
se x sese^ xx^ ++^ tantan^ xx dx
=
Z (^) se 2 x + se x tan x se x + tan x dx =
Z
du u = ln juj + C = ln j se x + tan xj + C : (31)
To integrate se 3 x, we use integration by parts with
u = se x; dv = se 2 x dx (32)
whi h yields du = se x tan x dx; v = tan x: (33)
It follows that (^) Z se 3 x dx = se x tan x
Z
se x tan 2 x dx: (34)
The identities
sin A os B = 1 2
[sin(A B ) + sin(A + B )℄; (43)
sin A sin B = 1 2
[ os (A B ) os(A + B )℄; (44)
os A os B = 12 [ os (A B ) + os(A + B )℄; (45)
whi h an b e derived from the identities for osines and sines of sums and di eren es, an b e used to onvert these integrands into mu h simpler ones involving only a sine or osine, in whi h ase the rules for integrating these fun tions an b e employed.
Example 5 Prove that for any nonnegative integers m and n, Z (^)
sin mx os nx dx = 0 : (46)
Solution Using the identity
sin A os B =
2 [sin(A^ ^ B^ )^ +^ sin(A^ +^ B^ )℄;^ (47)
we obtain (^) Z (^)
sin mx os nx dx =
Z
sin(m n)x + sin(m + n)x dx: (48)
Sin e sin x is an o dd fun tion, it follows that sin k x is o dd for any integer k. Sin e the integral of any o dd fun tion over an interval of the form [ a; a℄ is equal to 0, we must have Z (^)
sin(m n)x dx = 0 ;
Z
sin(m + n)x dx = 0 ; (49)
and therefore we must have (^) Z (^)
sin mx os nx dx = 0 : (50)
2
By now it should b e lear that there are a great many integrals that annot b e evaluated using the Fundamental Theorem of Cal ulus without some ingenuity. Although we have greatly expanded the range of integrands for whi h integration rules an b e develop ed, there are still many integrands for whi h no su h rules exist, and therefore the de nition of the de nite integral must b e used instead.
