Calculus II: Integration of trigonometric functions
1. Of the form: ∫ sino(x)
cosr (x) dx or
∫ sinr(x)
coso (x) dx
o stands for odd integer, and r stands for real.
The technique is:
1. Write sino(x) = sine(x) . sin(x)
(e stands for even).
2. Using the equality cos2 + sin2 = 1
we express sine(x) in function of cos (x)
sine(x) = [1 - cos2(x)]e/2
Therefore
∫ sino(x) cosr (x) dx =
∫ sine (x) sin(x) cosr(x) dx =
∫ [1 - cos2(x)]e/2
cosr(x) sin(x) dx
∫ sino(x) cosr (x) dx =
∫ [1 - cos2(x)]e/2
cosr(x) sin(x) dx
Example 1
∫ sin3(x)
cos2 (x) dx =
∫ [1 - cos2(x)]2/2
cos2(x) sin(x) dx =
∫ [cos2(x) - cos4(x)]
sin(x) dx =
∫cos2(x) sin(x) dx -
∫cos4(x) sin(x) dx
Let
cos(x) = u
so du = - sin(x) dx ,
we obtain:
- ∫u2 du + ∫u4du =
- (1/3)u3 + (1/5)u5 =
- (1/3) cos3(x) + (1/5) cos5(x) + cst
∫ sin3(x) cos2 (x) dx =
- (1/3) cos3(x) + (1/5) cos5(x) + cst
2. Of the form: ∫ sine(x)
cose (x) dx or
∫ sine(x) dx or
∫ cose (x) dx
e stands for even
In these kinds, we use the three following identities:
1. cos2(x) = (1 + cos(2x))/2
2. sin2(x) = (1 - cos(2x))/2
3. sin(2x) = 2 sin(x) cos(x)
Example
I = ∫ sin4(x) cos2(x) dx
We have:
sin4(x) = (1 - cos(2x))2/4
cos2(x) = (1 + cos(2x))/2
Therefore
sin4(x) cos2(x) =
(1/8) (1 - cos(2x))2 .(1 + cos(2x)) =
(1/8) (1 - cos(2x)) . (1 - cos(2x)) (1 + cos(2x)) =
(1/8) (1 - cos(2x)) . (1 - cos2(2x)) =
(1/8) [1 - cos2(2x) -
cos(2x) + cos3(2x)]
=
(1/8) [cos3(2x) - cos2(2x) -
cos(2x) + 1]
Therefore
I = (1/8) [∫ cos3(2x) dx -
∫ cos2(2x) dx -
∫ cos(2x) dx +
∫1 dx]
1. First term:
cos3(2x) = cos2(2x) cos(2x) =
(1 - sin2(2x)) cos(2x) =
cos(2x) - sin2(2x)) cos(2x)
∫ cos3(2x) dx =
∫ cos(2x) dx - ∫ sin2(2x)) cos(2x) dx =
(1/2)∫ cos(2x) d(2x) - (1/2)∫ sin2(2x)) dsin(2x) =
(1/2) sin(2x) - (1/2) (1/3) sin3(2x)).
∫ cos3(2x) dx = cos2(2x) cos(2x) =
(1/2)∫ cos(2x) d(2x) - (1/2) (1/3) sin3(2x))
2. Second term:
cos2(2x) = (1 + cos(4x))/2
(1/2)∫ (1 + cos(4x)) dx = (1/2) [∫ (1 + cos(4x))] dx =
(1/2) [∫dx + ∫cos(4x))dx = (1/2) [x + (1/4)sin(4x)]
(1/2)∫ (1 + cos(4x)) dx = (1/2) [x + (1/4)sin(4x)]
3. Third term::
∫ cos(2x) dx = (1/2) sin(2x)
4 Fourth term::
∫1 dx = x
Therefore
I = (1/8) [(1/2)sin(2x) - (1/2) (1/3) sin3(2x)) -
(1/2) (x + (1/4)sin(4x)) - (1/2) sin(2x) + x ] + cst.
= (1/16) [sin(2x) - (1/3) sin3(2x)) -
x - (1/4)sin(4x) - sin(2x) + 2x ] + cst =
(1/16) [ - (1/3) sin3(2x)) - (1/4)sin(4x) + x] + cst
∫ sin4(x) cos2(x) dx =
(1/16) [ - (1/3) sin3(2x)) - (1/4)sin(4x) + x] + cst
3. Of the form: ∫ tanr(x)
sece (x) dx or
∫ cotgr(x)
cosece (x) dx
r stands for real, and e stands for even
The steps are:
1. We start to transform sece (x) into
sece - 2 (x) . sec2 (x)
2. Using sec2(x) = 1 + tan2(x), we express
sece - 2 (x) in a function of tan(x)
3. For the second kind of integral, we use
cosec2(x) = 1 + cotg2(x)
instead.
Example
∫ tanr(x) sec4(x) dx =
∫ tanr(x) sec2(x) sec2(x) dx =
∫ tanr(x) [1 + tan2(x)] sec2(x) dx =
∫ tanr(x) [1 + tan2(x)] sec2(x) dx =
= ∫ tanr(x) sec2(x) dx +
∫ tanr+2(x) sec2(x) dx
Let u = tan(x), so du = sec2(x) dx
=
∫ ur du +
∫ ur+2 du =
(1/(r + 1))ur+1 + (1/(r + 2))ur+2 =
(1/(r + 1)) tanr+1(x) + (1/(r + 2)) tanr+2(x)
+ cst.
∫ tanr(x) sec4(x) dx
=
(1/(r + 1)) tanr+1(x) + (1/(r + 2)) tanr+2(x)
+ cst
4. Of the form: ∫ tano(x)
secr(x) dx or
∫ cotgo(x)
cosecr (x) dx
r stands for real, and o stands for odd
The steps are:
1. We start to transform tano(x) secr(x) into
tano-1(x) secr-1(x) . tan(x) sec(x)
2. Using sec2(x) = 1 + tan2(x), we express then
tano - 1(x) in a function of sec(x)
3. For the second kind of integral, we use
cosec2(x) = 1 + cotg2(x)
instead.
Example
∫ tan3(x) secr(x) dx =
∫ tan2(x) secr-1(x) tan(x) sec(x) dx =
∫ [sec2(x) - 1] secr-1(x) tan(x) sec(x) dx =
=
∫ sec2(x) . secr-1(x) tan(x) sec(x) dx
- ∫ secr-1(x) . tan(x) sec(x) dx
=
∫ secr+1(x) tan(x) sec(x) dx
- ∫ secr-1(x) . tan(x) sec(x) dx =
Let u = sec(x), so du = sec(x) tan(x) dx
=
∫ ur+1 du
- ∫ ur-1 du =
(1/(r+2))ur+2 - (1/(r)) ur
=
(1/(r+2)) secr+2(x) - (1/(r)) secr(x) + cst.
∫ tan3(x) secr(x) dx =
(1/(r+2)) secr+2(x) - (1/(r)) secr(x) + cst.
5. Of the form: ∫ tane(x)
seco(x) dx or
∫ cotge(x)
coseco (x) dx
e stands for even, and o stands for odd
The technique is integration by parts.
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