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Trigonometric Integrals
R
The book gives ::::
lots of rules/strategies for Trigonometric Integrals
(e.g. tan2 x cos x dx).
You are welcome to read them but do NOT memorize. Use logic
instead, as done below!
:::::
• Recall the following natural pairings (in terms of derivatives):
sin ↔ cos
tan ↔ sec
cot ↔ csc
If the integrand does not contain natural pairings, then rewrite it so that it does.
2x
sin2 x
sin2 x
or tan2 x cos x = tan2 x sec1 x = tan
.
Eg:
tan2 x cos x = cos
2 x cos x = cos x
sec x
• Next try a u-du substitution. There will be two naturals choice for u.
◦ If the integrand involves sine and/or cosine, then try u = cos x or u = sin x.
◦ If the integrand involves tangent and/or secant, then try u = tan x or u = sec x.
◦ If the integrand involves cotangent and/or cosecant, then try u = cot x or u = csc x.
• First
isolate off your du (box him and don’t touch him anymore)
and
:::::
then
try
to
express
what’s
left
in
terms
of
only
u.
:::::
The following will be helpful (and you must memorize them):
1 − cos(2x)
1 + cos(2x)
and
sin2 x =
(half-angle ( 12 ∠) formulas)
cos2 x =
2
2
cos(2x) = cos2 x − sin2 x
and
sin(2x) = 2 sin x cos x
(double-angle (2∠) formulas)
cos2 x sin2 x
1
+
=
=⇒
1 + tan2 x = sec2 x
2
2
cos x cos x
cos2 x
cos2 x sin2 x
1
cos2 x + sin2 x = 1
=⇒
+
=
=⇒
cot2 x + 1 = csc2 x
2
2
2
sin
x
sin
x
sin
x
R
R
R
I. For integrands of the form: sin (αx) cos (βx) dx or sin (αx) sin (βx) dx or cos (αx) cos (βx) dx
where α, β ∈ R and α::::::
6= β (see Example 5) use the trig identities (these 3 you do not have to memorize)
(1) sin (A) cos (B) = 21 [sin(A − B) + sin(A + B)]
cos2 x + sin2 x = 1
=⇒
(2) sin (A) sin (B) =
(3) cos (A) cos (B) =
1
[cos(A
2
1
[cos(A
2
− B) − cos(A + B)]
− B) + cos(A + B)] .
u=cos x
Example
4. sin4 x dx. u-du sub does not work (why? e.g.: sin4 x dx = − sin3 x [− sin x dx]).
R
For sinn x cosm x dx, with both
m, n ∈ {0, 2, 4, 6, . . .}, use the half-angle formulas.
:::::
2
Z
Z
Z Z
( 12 ∠)
A
A 1
1 − cos(2x)
4
2
2
dx =
[1 − 2 cos(2x) + cos2 (2x)] dx
sin x dx = [sin x] dx =
2
4
Z ( 21 ∠) 1
1 + cos(4x)
=
1 − 2 cos(2x) +
dx
4
2
Z
Z
Z
Z
1
1
1 1
1 1
=
dx −
2 cos(2x) dx + ·
dx + ·
cos(4x) dx
4
4
4 2
4 2
Z
Z
Z
1 1
1
1
1
1
·
=
1+
dx − · cos(2x) [2dx] +
·
cos(4x) [4dx]
4
2
4
4 2
4
1
1
3
= x − sin(2x) +
sin(4x) + C
8
4
32
R
R
R
to here is fine now but in trig-sub we’ll want all angles to be (1x) so let’s do it . . . note: sin (4x) = sin (2 (2x))
3
1
1
x − (2 sin x cos x) +
[(2) sin(2x) cos(2x)] + C
8
4
32
1
1 (2∠) 3
= x − (sin x · cos x) +
(2) (2 sin x · cos x) cos2 x − sin2 x + C
8
2
32
Good enough for now.
R
Example 5. sin (2x) cos (3x) dx. Hint: use (1) above with A = 2x, B = 3x. Soln is on bottom of page 4.
(2∠)
=
17.01.02 (yr.mn.dy)
Page 1 of 4
Techniques of Integration
Trigonometric Integrals
Z
sin7 x dx
Example 1.
u = cos x
du = − sin x dx
try 1:
First
box off your du:
:::::
u = cos x
Z
u = sin x
du = cos x dx
or
::
so
du = − sin x dx
Z
7
6 sin x dx = −
sin x − sin x dx .
Then
try to express what’s left, i.e. sin6 x , in terms of only u, i.e., in terms of only cos x:
:::::
sin6 x = (sin x)6 = sin2 x
3
= 1 − cos2 x
3
.
So:
Z
Z
7
sin6 x − sin x dx
Z h
3 i
=−
sin2 x
− sin x dx
Z h
3 i
2
=−
− sin x dx
1 − cos x
sin x dx = −
, did it.
(∗Ex.1 for u = cos x)
in Examples 1–3, once you get the the ∗-ed ::::::
clever:::::
way :::
to :::::
look :::
at ::::
the ::::::::
integral:::
so::::
can::::
do :::::
u-du::::
sub,
it’s smooth sailing from here on out!
Z h
3 i
=−
1 − u2
du
Z
A
=
u6 − 3u4 + 3u2 − 1 du
u7 3u5 3u3
−
+
−u+C
7
5
3
cos7 x 3 cos5 x
=
−
+ cos3 x − cos x + C
7
5
=
try 2:
u = sin x
so
du = cos x dx
First
box off your du:
:::::
Z
7
sin x dx =
Z
[???] would want here cos x dx
.
The substitution with u = sin x doesn’t work because there is no cos x in the integrand to help
make the needed du.
17.01.02 (yr.mn.dy)
///////////
Page 2 of 4
Techniques of Integration
Trigonometric Integrals
Z
Example 2.
sin4 (17x) cos3 (17x) dx
u = cos(17x)
du = −17 sin(17x) dx
try 1:
u = sin(17x)
du = 17 cos(17x) dx
or
::
u = cos (17x)
so
du = −17 sin (17x) dx
First
box off your du:
:::::
Z
Z
3
−1
4
3
sin (17x) cos (17x) dx =
cos (17x) sin3 (17x) −17 sin(17x) dx .
17
3
Then
try to express what’s left, i.e. cos (17x) sin3 (17x) , in terms of only u, i.e., in terms of only
:::::
cos (17x) :
cos3 (17x) sin3 (17x) = cos3 (17x) sin2 (17x) sin(17x) = cos3 (17x) 1 − cos2 (17x) sin(17x) .
Cannot do, there is an “extra sin”
try 2:
///////////
u = sin (17x)
so
du = 17 cos (17x) dx
First
box off your du:
:::::
Z
Z
4
1
4
3
sin (17x) cos (17x) dx =
sin (17x) cos2 (17x) 17 cos(17x) dx .
17
4
Then
try to express what’s left, i.e. sin (17x) cos2 (17x) , in terms of only u, i.e., in terms of only
:::::
sin(17x):
sin4 (17x) cos2 (17x) = sin4 (17x) 1 − sin2 (17x)
So: Z
, did it.
sin4 (17x) cos3 (17x) dx
Z
4
1
sin (17x) cos2 (17x) 17 cos(17x) dx
=
17
Z
4
1
=
sin (17x) 1 − sin2 (17x) 17 cos(17x) dx
(∗Ex.2 for u = sin(17x))
17
Z
4
1
=
u 1 − u2
du
17
Z
A 1
=
(u4 − u6 ) du
17
1 u5 u7
1 sin5 (17x) sin7 (17x)
=
−
+C =
−
+C .
17 5
7
17
5
7
17.01.02 (yr.mn.dy)
Page 3 of 4
Techniques of Integration
Trigonometric Integrals
Z
Example 3.
tan θ sec4 θ dθ
try 1:
u = tan θ
du = sec2 θ dθ
so:
u = tan θ
so
or
::
u = sec θ
du = sec θ tan θ dθ
du = sec2 θ dθ
First box off your du:
:::::
Z
Z
4
tan θ sec θ dθ =
tan θ sec2 θ sec2 θ dθ .
Then
try to express what’s left, i.e. [tan θ sec2 θ], in terms of only u, i.e. in terms of only tan θ:
:::::
tan θ sec2 θ = tan θ 1 + tan2 θ
did it.
So:
Z
Z
4
tan θ sec θ dθ =
,
tan θ sec2 θ sec2 θ dθ
Z
tan θ 1 + tan2 θ sec2 θ dθ
(∗Ex.3 for u = tan θ)
Z
Z
A
u2 u4
tan2 θ tan4 θ
2
(u + u3 ) du =
+
+C =
+
+C
=
u(1 + u ) du =
2
4
2
4
=
try 2:
u = sec θ
so
du = sec θ tan θ dθ
First
box off your du:
:::::
Z
4
Z
tan θ sec θ dθ =
3 sec θ sec θ tan θ dθ .
,
Then
try to express what’s left, i.e. [sec3 θ], in terms of only u, i.e. in terms of only sec θ.
easy
:::::
So:
Z
Z
3 4
tan θ sec θ dθ =
(∗Ex.3 for u = sec θ)
sec θ sec θ tan θ dθ
Z
3
u4
sec4 θ
=
u
du =
+K =
+K
4
4
Our answers from try 1 and try 2 should be the same . . . and they are since:
2
2
A 1
T 1
sec4 θ
+K =
sec2 θ + K =
1 + tan2 θ + K
4
4
4
2
A tan θ
A 1
tan4 θ
1
2
4
1 + 2 tan θ + tan θ + K =
+
+ K+
.
=
4
2
4
4
So
1
C=K+ .
4
Now onto Example 4 (on page 1) and below is the Solution to Example 5. Using trig identities:
first (1) sin (A) cos (B) = 21 [sin(A − B) + sin(A + B)] and then last cos(−x) = cos x.
Z
Z
Z
1
(1) 1
sin (2x) cos (3x) dx =
[sin (2x − 3x) + sin (2x + 3x)] dx =
[sin (−x) + sin (5x)] dx
2
2
Z
Z
1
1
1
=
(−1)
sin (−1x) (−1 dx) +
sin (5x) (5 dx)
2
2
5
cos (−x) cos(5x)
cos x cos(5x)
−
+C =
−
+C .
=
2
10
2
10
17.01.02 (yr.mn.dy)
Page 4 of 4
Techniques of Integration