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      Calculus II

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© The scientific sentence. 2010

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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