Trigonometric Integrals and Substitution

Trigonometric Integrals
Howdoyouintegrateanexpressionlike sinnx cosmxdx?(n =0, 1, 2... and m =0, 1, 2,...)
Wealreadyknowthat:
sin xdx = ?cos x + c and cos xdx = sin x + c
Method A
Supposeeither n or m isodd.
Example 1. sin3x cos2xdx.
Ourstrategyistouse sin2x + cos2x =1 torewriteourintegralintheform:
sin3x cos2xdx = f(cosx) sinxdx
Indeed, ?? ?
sin3x cos2xdx = sin2x cos2x sin xdx = (1 ? cos2x)cos2x sin xdx
Next,usethesubstitution
u = cos x and du = ?sin xdx Then, ??
(1 ? cos2x) cos2x sin xdx = (1 ? u2)u2(?du)
11 11
=(?u2 + u4)du = ?3u3 +5u5 + c = ?3 cos3u + 5 cos5x + c
Example 2. ?? ?
cos3xdx = f(sin x) cos xdx = (1 ? sin2x) cos xdx
Again,useasubstitution,namely u = sin x and du = cos xdx
u3 sin3x
cos3xdx = (1 ? u2)du = u ? + c = sin x ? + c
33
1
? ?
?
? ? ?
?
? ?
Lecture 26 18.01 Fall 2006
Method B
Thismethodrequires both m and n to be even. It requires double-angle formulae such as
1 + cos2x
2
cosx = 2
(Recallthat cos2x = cos2 x ? sin2 x = cos2 x ? (1 ? sin2 x) = 2cos2 x ? 1) Integratinggetsus ??
1 + cos2xx sin(2x)
cos2 xdx = dx =+ + c
2 24 Wefollowasimilarprocessforintegrating sin2x.
1 ? cos(2x)sin2 x = 2
1 ? cos(2x) x sin(2x)
sin2xdx = dx =+ c
22 ? 4
The full strategy for these types of problems is to keep applying Method B until you can apply MethodA(whenoneof m or n isodd).
Example 3. sin2x cos2xdx.
ApplyingMethodBtwiceyields
???? ??? ?
1 ? cos2x 1 + cos2x 11
22 dx =4 ? 4cos22x dx
11 11
=4 ? 8(1 + cos4x) dx =8x ? 32 sin4x + c
There is a shortcut for Example 3. Because sin2x = 2 sin x cos x,
? ???2 ?
sin2x cos2xdx = 1 sin2x dx =1 1 ? cos4xdx = sameasabove
2 42
Thenextfamilyoftrigintegrals,whichwe’llstarttoday,butwillnotfinishis:
secnx tanmxdx where n =0, 1, 2,... and m =0, 1, 2,...
Rememberthat
sec2 x = 1+tan2 x
whichwedoublecheckbywriting
1 sin2 x cos2 x + sin2 x
=1+ =
cos2 x cos2 x cos3 x
sec2 xdx = tan x + c sec x tan xdx = sec x + c
2
? ?
? ? ?
?
?
? ? ?
? ?
? ?
?
?
Lecture 26 18.01 Fall 2006
Tocalculatetheintegralof tan x,write
sin x
tan xdx = dx
cos x
Let u = cos x and du = ?sin xdx,then
sin x du
tan xdx = cos x dx = ? u = ?ln(u)+ c
tan xdx = ?ln(cos x)+ c
(We’llfigureoutwhat sec xdx islater.)
Now, let’s see what happens when you have an even power of secant. (The case n even.)
sec4xdx = f(tanx) sec2xdx = (1 + tan2x) sec2xdx
Makethefollowingsubstitution:
u = tan x
and
du = sec2xdx
u3 tan3 x
sec4xdx = (1+ u2)du = u ++ c = tan x ++ c
33
Whathappenswhenyouhaveaoddpowerof tan? (The case m odd.)
tan3x sec xdx = f(sec x) d(sec x)
= (sec2x ? 1) sec x tan xdx
(Rememberthat sec2 x ? 1 = tan2 x.) Usesubstitution:
u = sec x
and
du = sec x tan xdx
Then, ??
u3 sec3 x
tan3x secxdx =(u2 ? 1)du =3 ? u + c =3 ? sec x + c
Wecarryoutonefinalcase: n =1,m =0
sec xdx = ln (tan x + sec x)+ c
3
? ? ? ?
?
?
Lecture 26 18.01 Fall 2006
We get the answer by “advanced guessing,” i.e., “knowing the answer ahead of time.”
sec x + tan x sec2 x + sec x tan x sec x dx = sec x dx = dx sec x + tan x tan x + sec x
Make the following substitutions:
u = tan x + sec x
and
du = (sec2 x + sec x tan x) dx
This gives ? ?
du sec x dx = = ln(u) + c = ln(tan x + sec x)