Integrals of Trigonometric Functions

Six Important Integrals Involving Trig Functions

In the previous discussion we gave the derivatives of the six basic trig functions.  Since integrals are antiderivatives, we can read the table backwards and arrive at six basic trig integrals.  We list the table again, adjusting with  a "+ C".

 f (x) sin x + C cos x cos x + C -sin x tan x + C sec2 x sec x + C sec x tan x csc x + C -csc x cot x cot x + C -csc2 x

Example

Find

sec(2x)tan(2x) dx

Solution

This does not exactly fit the table, however with a u-substitution we can adjust.  Let

u  =  2x        du  =  2dx        dx  =  1/2 du

Substituting, gives

1/2 sec u  tan u du

Now we can recognize this as the derivative of sec u, hence the integral is

1/2 sec u  +  C     =    1/2 sec(2x)  +  C

Exercise

Evaluate the following integrals  (Hold your mouse on the yellow rectangle for the answer)

A.   sin x sec2(cos x) dx

B.   x csc(x2)cot(x2) dx

The Integrals of the Six Basic Trig Functions

Now that we know what integrals produce the six basic trig functions, we would like to know how to find the integrals of the six basic trig functions.  While sin x and cos x come straight from the antiderivatives, the other four are more difficult to find.  We begin with tan x.

tan x dx

The trick to find this integral is to use the sin over cos definition of the tan.  We have

Now we can use u-substitution.  Let

u  =  cos x    du  =  -sin x dx

This gives us

-1/u du  =  -ln|u| + C  =  -ln|cos x| + C

sec x dx

This one is even trickier.  We multiply by the term

sec x + tan x

sec x + tan x

I know, that is just what you were thinking of.  We have

and behold!  The numerator is the derivative of the denominator.  This leads us to the substitution

u  =  sec x + tan x        du  =  (sec x tan x + sec2 x) dx

Now substitute to get

1/u du  =  ln|u| + C  =  ln|sec x + tan x| + C

csc x dx

This integral is most easily accomplished by using the trig identity

csc x  =  sec(p/2 - x)

We have

csc x dx  =  sec(p/2 - x) dx

Let

u  =  p/2 - x    du  =  -dx

This gives us

-sec(u) du  =  - ln|sec u + tan u| + C

Now resubstitute to get

-ln|sec(p/2 - x) + tan(p/2 - x)| + C

and use the trig identities again to get

-ln|csc x + cot x| + C

Exercise

Show that

cot x dx  =  ln|sin x| + C

We summarize with the following table

 f(x) sin x -cos x + C cos x sin x + C tan x -ln|cos x| + C sec x ln|sec x + tan x| + C csc x -ln|csc x + cot x| + C cot x ln|sin x| + C

Application

An adult on a tread mill has a respiratory cycle with air flow approximately

pt
v  =  |1.4sin         |
2

where t is the time in seconds and v is the velocity in liters per second.  Find the volume of air inhaled during one cycle.

Solution

The integral of the velocity will give the total volume inhaled.  One cycle goes from 0 to 2 seconds.  We have

Let

u  =  p t/2          du  =  p/2 dt        dt  =  2/p du

when

t  =  0    u  =  0

and when

t  =  2    u  =  p

We get

The volume of air in one cycle is about 1.78 liters.

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