Week4
Math 112 - 018 Rahman
8.2 Trigonometric Integration
Lets look at a few examples first and then we’ll develop a general strategy.
Sines and Cosines.
R
(1) Consider cos3 xdx.
Solution: We recall the identity: cos2 = 1 − sin2 x. Lets see if we can use
this to simplify the problem.
Z
Z
3
Z
2
cos x[1 − sin x]dx =
cos xdx =
Z
cos xdx −
sin2 x cos xdx.
Now, the first integral is easy and the second integral we solve via u-sub
where u = sin x ⇒ du = cos xdx.
Z
(2)
Z
cos3 xdx = sin x −
Z
1
1
u2 du = sin x − u3 + C = sin x − sin3 x + C
3
3
sin5 x cos2 xdx.
Solution: Lets use the same strategy as above, except this time on sin x.
R
5
2
Z
sin x cos xdx =
2
2
2
Z
(sin x) sin x cos xdx =
(1 − cos2 x)2 cos2 x sin xdx.
We can go straight to u-sub with u = cos x ⇒ du = − sin x,
Z
sin5 x cos2 xdx = −
(3)
sin2 xdx
Solution: For this problem if we used the identity we used for the past
two problems we would be going in circles, so we use another identity - the
double angle formula: cos 2x = 1 − 2 sin2 x,
0
π
sin2 xdx =
0
Z
1
2
1
1
2
1
(1−u2 )2 u2 du = − u3 + u5 − u7 +C = − cos3 x+ cos5 x− cos7 x+C
3
5
7
3
5
7
Rπ
Z
(4)
Z
1
2
Z
π
(1 − cos 2x)dx =
0
π
1
1
π
x − sin 2x
=
2
2
2
0
sin4 xdx
Solution: This is similar to the above problem,
R
2
Z
1
1
sin xdx =
(1 − cos 2x) dx =
(1 − 2 cos 2x + cos2 2x)dx
2
4
Z 1
1
1 3
1
=
1 − 2 cos 2x + (1 + cos 4x) dx =
x − sin 2x + sin 4x + C
4
2
4 2
8
4
Z 1
Strategies for
sinm x cosn xdx.
R
(1) If the power of the cosine term is odd (i.e. n = 2k + 1), save one cosine
factor and use cos2 x = 1 − sin2 x,
Z
m
sin x cos
2k+1
Z
sinm x(cos2 x)k cos xdx
Z
sinm x(1 − sin2 x)k cos xdx.
dx =
=
(1)
Then substitute u = sin x ⇒ du = cos x.
(2) If the power of the sine term is odd (i.e. m = 2k + 1), save one sine factor
and use sin2 x = 1 − cos2 x,
Z
sin2k+1 cosn xdx =
Z
(sin2 x)k cosn xdx =
Z
(1 − cos2 x)k cosn x sin xdx.
(2)
Then substitute u = cos x ⇒ du = − sin x.
(3) If the powers of both sine and cosine are even, use the double-angle formulas:
sin2 x =
1
(1 − cos 2x)
2
cos2 x =
1
(1 + cos 2x)
2
sin x cos x =
1
sin 2x.
2
Tangents and Secants.
(1)
tan6 x sec4 xdx.
Solution: We recall the identity sec2 x = 1 + tan2 x, and see where this
takes us
R
Z
tan6 x sec4 xdx =
Z
tan6 x(1 + tan2 x) sec2 xdx.
Then we substitute u = tan x ⇒ du = sec2 xdx, then
Z
tan6 x sec4 xdx =
(2)
Z
5
Z
u6 (1 + u2 )du =
1 7 1 9
1
1
u + u + C = tan7 x + tan9 x + C
7
9
7
9
tan5 θ sec7 θdθ.
Solution: Here lets try using the other identity: tan2 x = sec2 x − 1,
R
7
Z
tan θ sec θdθ =
4
6
Z
tan θ sec θ sec θ tan θdθ =
(sec2 θ−1)2 sec6 θ sec θ tan θdθ.
We employ the u-sub u = sec θ ⇒ du = sec θ tan θdθ,
Z
5
7
tan θ sec θdθ =
Z
(u2 −1)2 u6 du =
1 11 2 9 1 7
1
2
1
u − u + u +C =
sec11 x− sec9 x+ sec7 x+C
11
9
7
11
9
7
R
Strategies for tanm x secn xdx.
(1) If the power of the secant term is even (i.e. n = 2k, k ≥ 2), save a factor
of sec2 x and use sec2 x = 1 + tan2 x,
Z
tanm x sec2k xdx =
Z
Z
=
tanm x(sec2 x)k−1 sec2 xdx
tanm x(1 + tan2 x)k−1 sec2 xdx.
(3)
Then substitute u = tan x ⇒ du = sec2 xdx.
(2) If the power of the tangent term is odd (i.e. m = 2k + 1), save a factor of
sec x tan x and use tan2 x = sec2 x − 1,
Z
tan
2k+1
Z
n
x sec xdx =
Z
=
(tan2 x)k secn−1 x sec x tan xdx
(sec2 x − 1)k secn−1 x sec x tan xdx.
(4)
Then substitute u = sec x ⇒ du = sec x tan xdx
Useful Integrals.
These integrals are also pretty easy to derive if you forget them,
Z
tan xdx = − ln | cos x| + C = ln | sec x| + C.
(5)
Z
sec xdx = ln | sec x + tan x| + C.
(1)
tan3 xdx.
Solution: We use the identity tan2 x = sec2 x − 1,
R
Z
(2)
Z
3
(6)
tan3 xdx =
Z
tan x(sec2 x − 1)dx =
1
tan2 x − ln | sec x| + C.
2
sec3 xdx.
Solution: We integrate by parts with u = sec x ⇒ du = sec x tan xdx and
dv = sec2 x ⇒ v = tan x, then
R
Z
Z
sec x tan xdx = sec x tan x − sec x(sec2 x − 1)dx
Z
Z
Z
= sec x tan x − sec3 xdx + sec xdx = sec x tan x − sec3 xdx + ln | sec x + tan x|
Z
1
⇒ sec3 xdx = [sec x tan x + ln | sec x + tan x|] + C.
2
sec xdx = sec x tan x −
2
Useful identities that you probably wont have to use that much.
1
sin a cos b = [sin(a − b) + sin(a + b)]
2
1
sin a sin b = [cos(a − b) − cos(a + b)]
2
1
cos a cos b = [cos(a − b) + cos(a + b)].
2
R
(1) sin 4x cos 5xdx.
Solution: We use the first identity to get,
Z
Z
1
1
1
sin 4x cos 5xdx =
(sin(−x) + sin 9x)dx = cos x −
cos 9x + C.
2
2
18
One can also do this problem by parts, which is actually the preferred
method, but a little extra knowledge never hurt anyone.
(7)
(8)
(9)