Trigonometric
Integral
Examples…
Sine
or
Cosine
to
odd
powers…
5
∫ cos
xdx
2
∫ (1− sin x ) cos xdx
= ∫ (1− u ) du
= ∫ (1−2u + u )du
2
=
2
2
2
4
= u − 23 u3 + 15 u5 + C
= sin x − 23 sin3 x + 15 sin5 x + C
€
Sine
or
Cosine
to
even
powers…
4
∫ sin
xdx
2
2
1
2
1
2
1
4
1
2
1
4
1
4
1
2
1 1
4 2
3
8
1
2
1
8
3
8
2
∫ (sin x ) dx
= ∫ ( − cos(2x )) dx
= ∫ ( − cos(2x ) + cos (2x ))dx
= ∫ ( − cos(2x ) + ( + cos(4x )))dx
= ∫ ( − cos(2x ) + cos(4x ))dx
= x − sin(2x ) + sin(4x ) + C
=
1
4
2
1
2
1
32
€
Tangent
to
powers…
5
∫ tan
xdx
∫ (sec x − 1) tan xdx
= ∫ (tan x sec x )dx − ∫ tan xdx
= ∫ u du − ∫ (sec x − 1) tan xdx
= ∫ u du − ∫ tan x sec xdx + ∫ tan xdx = ∫ u du − ∫ udu + ∫ tan xdx
2
=
3
3
3
3
2
3
2
2
3
= 14 u4 − 12 u2 − lncos x + C
€
= 14 tan4 x − 12 tan2 x − ln cos x + C
4
∫ tan
∫ (sec x − 1) tan xdx
= ∫ tan x sec xdx − ∫ tan
= ∫ u du − ∫ (sec x − 1)dx
xdx
2
=
2
2
2
2
2
xdx
2
= 13 u3 − tan x + x + C
= 13 tan3 x − tan x + x + C
€
Secant
to
even
powers…
∫ sec
6
2
∫ (1+ tan x ) sec xdx
= ∫ (1+2tan x + tan x ) sec xdx
= ∫ (sec x +2tan x sec x + tan x sec x )dx
= tan x + ∫ (2u + u )du
xdx
2
=
2
2
4
2
2
2
2
2
4
2
4
= tan x + 23 u3 + 15 u5 + C
= tan x + 23 tan3 x + 15 tan5 x + C
€
Secant
to
odd
powers
is
an
ugly
(I
mean
beautiful)
integration
by
parts
process.
∫ sec3 xdx First
we
use
integration
by
parts
with
u = sec x and dv = sec2 xdx ∫ sec
3
xdx
€
∫ sec x €tan xdx
= sec x tan x − ∫ sec x (sec x − 1)dx
= sec x tan x − ∫ sec xdx + ∫ sec xdx
= sec x tan x + ln sec x + tan x − ∫ sec
2
3
xdx
Which
gives:
€
3
2
= sec x tan x −
2 sec3 xdx = sec x tan x + ln sec x + tan x
∫
€
⇒
∫
sec3 xdx = 12 sec x tan x + 12 ln sec x + tan x + C