Download Lecture 5-1D: Integrating Trigonometric Functions

Section 5 – 1D:
Integrals involving the basic Trigonometric Functions
The integral of a function is the anti derivative of that function. If you know the derivative of a function
F(x) is f (x) then you also know that the integral of ∫ f (x) dx = F(x) . We know the derivatives or the
6 basic trigonometric functions. This allows us to state the derivatives of these 6 functions as an
integral.
Example 1
Example 2
Example 3
We know that
d
sin(x) = cos(x)
dx
We know that
d
sec(x) = sec(x)tan(x)
dx
We know that
d
tan(x) = sec 2 (x)
dx
so we know the integral
so we know the integral
so we know the integral
∫ cos(x) dx
must be sin(x) + C
∫ sec(x) tan(x) dx
must be sec(x) + C
2
∫ sec (x) dx
must be tan( x) + C
∫ cos(x) dx = sin(x) + C
∫ sec(x) tan(x) = sec(x) + C
2
∫ sec (x) = tan( x) + C
Example 4
Example 5
Example 6
We know that
d
cos(x) = − sin(x)
dx
We know that
d
csc(x) = − csc(x)cot(x)
dx
We know that
d
cot(x) = − csc 2 (x)
dx
so we know the integral
so we know the integral
∫ − sin(x) dx
must be cos(x) + C
so we know the integral
− csc(x) cot( x) dx
must be csc(x) + C
∫ − sin(x) dx = cos(x) + C
∫ − csc(x)cot(x) dx = csc(x) + C
2
∫ − csc (x) = cot( x) + C
2
∫ − csc (x) dx
must be cot(x) + C
Note: The first 3 integrals are commonly used in this course. The use of the last 3 is less common
but we state all 6 for completeness.
Examples 3 , 4 , and 5 above are all functions with a coefficient of negative 1.
∫ − sin(x) dx = cos(x) + C
∫ − csc(x)cot(x) dx = csc(x) + C
2
∫ − csc (x) = cot( x) + C
We will now develop techniques to integrate these functions with a coefficient of positive 1.
∫ sin(x) dx
Section 5 – 1D
∫ csc(x)cot(x) dx
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∫ csc (x)
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An Integral can be multiplied by a constant times its reciprocal
1
∫ f (x) dx = • ∫ a• f (x) dx
a
1
because • a = 1
a
so
∫ f (x) dx = − ∫ − f (x) dx
because − 1• −1 = 1
This rule that allows us to multiply the outside and inside of an integral by –1
∫ f (x) dx = − ∫ − f (x) dx
Example 7
Example 8
Example 9
d
cos(x) = − sin(x)
dx
Find
d
csc(x) = − csc(x)cot( x)
dx
Find
d
cot(x) = − csc 2 (x)
dx
Find
∫ sin(x) dx
∫ csc(x)cot(x) dx
2
∫ csc (x) dx
multiply inside and outside
by − 1 because − 1• −1 = 1
to get − ∫ − sin(x) dx
multiply inside and outside
by − 1 because − 1• −1 = 1
to get − ∫ − csc(x) cot( x) dx
multiply inside and outside
by − 1 because − 1• −1 = 1
∫ sin(x) dx = − cos(x) + C
∫ csc(x)cot(x) dx = − csc(x) + C
to get − ∫ − csc 2 (x) dx
2
∫ csc (x) dx = − cot(x) + C
These 3 integral rules above match the rules commonly listed for the basic trig integrals.
The listed Trigonometric integrals rules
The 6 rules for integrals shown below require the function to be exactly the Trigonometric
function shown with the variable be exactly x.
∫ cosx dx = sin x + C
∫ sec x tan x dx = sec x + C
2
∫ sec x dx = tan x + C
∫ sinx dx = − cos x + C
∫ csc x cot x dx = − csc x + C
2
∫ csc x dx = − cot x + C
THESE RULES CANNOT be used for functions like ∫ cos3x dx or ∫ sinx 2 dx or ∫ sec2 (x + 2) dx
because the variable is not exactly x.
Section 5 – 1D
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© 2015 Eitel
What's the difference between ∫ sin(x) dx and ∫ − sin(x) dx
The derivative of cos(x) yields an expression with a negative coefficient.
d
cos(x) = − sin(x)
dx
Since we are determining anti derivatives we now have a rule for the integral of ∫ − sin(x)
d
cos(x) = − sin(x)
dx
so
∫ − sin(x) dx = cos x + C
The use of the ∫ f (x) dx = − ∫ − f (x) dx rule for integrals
allows us to develop a rule for ∫ sin(x) dx
∫ sin(x) dx
multiply inside and outside by − 1 to get
− ∫ − sin(x) dx
= − cos(x) + C
When we start to create a list or table of integral rules for basic functions
you would think we would would use
∫ − sin(x) dx = cos x + C
as it is derived from an anti derivative
That is not the common choice. All tables of basic integrals list the function with a positive
coefficient. The student then memorizes that version and uses it as the basic rule.
This causes 3 of the integrals for basic trigonometric functions to be written as shown below
∫ sin(x) dx = − cos x + C
is listed as the basic integral instead of
∫ − sin(x) dx = cos x + C
2
∫ csc x dx = − cot x + C
is listed as the basic integral instead of
2
∫ − csc x dx = cot x + C
∫ csc x cot x dx = − csc x + C
is listed as the basic integral instead of
∫ − csc x cot x dx = csc x + C
Section 5 – 1D
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© 2015 Eitel
Useful Trigonometric Identities
sec x =
1
cos x
sin2 x + cos2 x = 1
csc x =
1
sin x
tan x =
tan2 x +1 = sec 2 x
1
cot x
cot 2 x +1 = csc 2 x
cos2x = cos2 x − sin 2 = 2 cos2 x − 1 = 1− 2 sin2 x
sin2x = 2 sin x cos x
Integrals Involving Trigonometric Identities
Example 1
∫
sin 2x
dx
cos x
the functions are not both in terms of x
but you can use the double angle rule
sin2x = 2 sin x cos x
Example 2
1− sin2 x
dx
∫
cos x
you can use sin2 x + cos2 x = 1
to derive 1− sin2 x = cos2 x
cos2 x
dx
cos x
2sinx cos x
dx
∫
cos x
∫
= 2∫ sin x dx
which integrates to
= − 2cos x + C
= ∫ cos x dx
which integrates to
= sinx + C
Example 3
Example 4
∫ sinx(csc x − 1) dx
∫ tanθ (cot θ − sec θ ) dθ
∫ (sin x csc x − sin x ) dx
∫ (tan θ cot θ − sec θ tanθ ) dθ
⎛
∫ ⎝ sin x
⎛
∫ ⎝ tanθ
1
⎞
− sin x dx
⎠
sin x
1
⎞
− sec x tan x dθ
⎠
cot x
= ∫ 1 dx − ∫ sin x dx
= ∫ 1 dθ − ∫ secθ tanθ dθ
which integrates to
= x + cos x + C
which integrates to
= θ − sec θ + C
Section 5 – 1D
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© 2015 Eitel