Chapter VII. D. Trigonometric Integrals

© 2014 M. Flashman 
   Integrals involving trigonometric functions are solved usually using some trigonometric identities and substitutions. In that sense they present nothing new as a problem in integration, merely one of recognizing a helpful substitution or identity. We consider a few of these methods by example and then leave others for your exploration in the exercises.
First we recall those integrals of trigonometric functions we have encountered already .

Integrals of Core Trigonometric Functions
$\int   \sin(x)dx = -\cos(x)+C$ $\int    \cos(x)dx = \sin(x)+C$
$\int  \tan(x)dx = ln(│\sec(x)│)+C$ $\int  \sec(x)dx = ln(│\sec(x)+\tan(x)│)+C$

Examples Involving the Sine and Cosine Functions.

Example VII.D.1 Find the following indefinite integrals:a. $\int \sin ^6(x) \cos(x) dx$     b. $\int \sin ^6(x) \cos ^3(x) dx$         c. $\int \cos ^3(x) dx$

Solutions: In each of these examples we use the substitution  $u = \sin(x)$  so that $du = \cos(x) dx$ and $cos^2(x) = 1 - \sin^2(x) = 1 - u^2$.

a. With the substitution this integral becomes  $\int \sin ^6(x) \cos(x) dx = \int   u ^6 du = u ^7/7 + C =1/7 \sin ^7(x) + C$

b. With the substitution this integral becomes   $\int \sin^6(x) \cos ^3(x) dx = \int   u ^6 (1 - u ^2)du = u ^7/7 - u ^9/9 + C = 1/7 \sin^7(x) - 1/9 \sin^9(x) + C$

c. With the substitution this integral becomes   $\int \cos ^3(x) dx = \int   (1 - u ^2)du = u - u ^3/3 + C = \sin(x) - 1/3 \sin ^3(x) + C$

Remark: As long as there is an odd power of either sine or cosine in this kind of integral, a substitution $u = \sin(x)$ or $u = \cos(x) $ will work in general.

Example VII.D.2 Find the following indefinite integrals :  a. $\int \sin ^2(x) dx$    b. $\int \sin ^4(x) cos ^2(x) dx$

Solutions: For each of these examples in which only even powers of the sine and cosine appear we use the trigonometric identities

$sin ^2(x) = 1/2 [ 1 - \cos(2x) ]$  and $\cos ^2(x) = 1/2 [ 1 + \cos(2x) ]$ .

a. With the identity this integral becomes $\int \sin ^2(x)dx = \int   1/2 [1 - \cos(2x)]dx = 1/2[x - 1/2 \sin(2x)] + C = 1/2 [x - \sin(x)\cos(x)] + C$.
b. With the identities this integral becomes    $\int \sin ^4(x) \cos ^2(x) dx = \int  1/8 [1 - \cos(2x)] 2  [1 + \cos(2x)] dx = 1/8 \int  [1-cos(2x)][1-cos ^2(2x)]dx =1/8 \int  [1-cos(2x)][sin ^2(2x)] dx.$
$= 1/8\int \sin ^2(2x) - \cos(2x)\sin ^2(2x) dx$.

We leave the completion of this problem as an exercise for the reader using part a) and the methods discussed in Example VII.D.1.
Examples Involving the Tangent  and Secant Functions.

Example VII.D.3 Find the following indefinite integrals:   a. $\int   \tan(x) \sec ^6(x) dx$ b. $\int   \tan ^3(x) dx$

Solutions: In each of these examples we use the substitution  $u = \tan(x)$; so that $du = \sec ^2(x) dx$ and $ \sec ^2(x) = 1 + \tan ^2(x) = 1 + u ^2$.

a. With the substitution this integral becomes   $\int   \tan(x) \sec ^6(x) dx = \int   u(1 + u ^2) ^2 du = u ^2/2 +2u^4/4 +u^6/6 + C = 1/2 \tan ^2(x) + 1/2 \tan ^4(x) + 1/6 \tan ^6(x) + C$.

b. First we notice that $\tan ^3(x) = \tan(x)(sec ^2(x) - 1) = \tan(x)sec ^2(x) - \tan (x)$.
$\int   \tan ^3(x) dx = \int   \tan(x)sec ^2(x) dx - \int   \tan (x) dx$

[ Use the substitution to evaluate the first of these integrals and you'll arrive at the result.]

                  $= 1/2 \tan ^2(x) + \ln (│\cos(x)│) + C$

Remark: As long as there is an even power of secant together with the tangent in this kind of integral, a substitution $u = \tan(x)$ will work in general.

Example VII.D.3 a. Revisited.

Alternate Solution: Let $u = \sec(x)$ and then $du = \sec(x) \tan(x) dx$.

With this substitution we have     $\int  \sec ^6(x) \tan(x)dx = \int   u ^5 du = u ^6/6 + C = 1/6 \sec ^6(x) + C$

Remark: So there are cases where using secant for the substitution is possible and more efficient than using the tangent. The exercises explore some other techniques used for integrating expressions using tangents and secants.
Other Examples Involving the Sine and Cosine Functions. 

As the previous examples have illustrated, there is a certain amount of choice and good fortune involved in being able to solve an integration problem for trigonometric functions. To illustrate further this quality of trigonometric integrals here are some examples using a different type of trigonometric identity.

 Example VII.D.4. Find the following indefinite integrals:    a. $\int \sin(3x) \cos(2x) dx$      b. $\int \sin(3x) \sin(2x) dx$

Solutions: In each of these integrals we use one of the identities relating products to sums of trigonometric functions.

             $\sin(A) \cos(B) = 1/2 [ \sin(A + B) + \sin(A - B) ]$

             $\sin(A) \sin(B) = 1/2 [ \cos(A - B) - \cos(A + B) ]$

a) Let $A = 3x$ and $B = 2x$ in the first identity and we obtain
     $\int \sin(3x) \cos(2x) dx = \int   1/2 [ \sin(5x) + \sin(x) ] dx = - 1/2 [ 1/5 \cos(5x) + \cos(x) ] + C$.

b) Let $A = 3x$ and $B = 2x$ in the second identity and we obtain

   $\int \sin(3x) \sin(2x) dx = \int   1/2 [ \cos(x) - \cos(5x) ] dx = 1/2 [ \sin(x) - 1/5 \sin(5x) ] + C.$

   Some definite integrals involving the trignometric functions can be simplified considerably by the symmetry properties of the trignometric functions.

Example VII.D.5. Use symmetry to show  $\int_{-π}^\pi \sin ^3(x) \cos(x) dx = 0 $.

Solution: Since $\sin(-x) = - \sin(x)$ and $\cos(-x) = \cos(x)$ for all x, we have that the integrand $f(x) = \sin ^3(x) \cos(x)$ has $f(-x) = - f(x)$.

Now we break the integral into pieces using the additive property of integrals, to see that
$\int _{-π}^\pi \sin^3 (x) \cos(x)dx$ = $\int _0^\pi \sin ^3(x) \cos(x)dx$ + $\int _{-π}^0 \sin ^3(x) \cos(x)dx$

The substitution $u = - x$ and $du = - dx$ on the second integral in

this sum gives
$\int_{-π}^0 \sin ^3(x) \cos(x) dx = \int _π ^0 \sin ^3(-u) \cos(-u) (- du)  = \int_π^0 \sin ^3(u) \cos(u) du = - \int _0^\pi \sin ^3(u) \cos(u) du$.

   Replacing u with x in the last definite integral gives us $\int_{-π}^0 \sin ^3(x) \cos(x) dx = -\int_0^\pi \sin ^3(x) \cos(x) dx$ so  $\int_{-π}^\pi \sin ^3(x) \cos(x) dx = 0$.
Use Evaluate to have SAGE graph
$\sin ^3(x) \cos(x)$

A graph of the integrand $\sin ^3(x) \cos(x)$ for the interval $[-π,π]$ illustrates the geometry of the symmetry here, as well as why the integral is zero.
Can you give a geometric interpretation and / or a motion interpretation in terms of differential equations for this integral being equal to zero?
 

Exercises VII.D

Find the following indefinite integrals:
  1. $\int \sin ^2(x) \cos(x) dx$

  2. $\int   \tan ^2(x) \sec ^2(x) dx$

  3. $\int \sin ^3(x) dx$

  4. $\int \cos ^2(x) \sin ^2(x) dx$

  5. $\int  \tan(x) \sec ^3(x) dx$

  6. $\int  \sec ^4(x) dx$

  7. $\int \sin ^2(x) \tan(x) dx$

  8. $\int  \tan ^5(x) dx$

  9. $\int \sin ^4(x) dx$

  10. $\int \sin ^3(x) \cos ^2(x) dx$


    For problems 11-16 find the indicated definite integrals.

  11.  $\int_0 ^{\pi /2}\sin ^3(x) \cos(x) dx$

  12. $\int_0^{\pi /4} \sec ^2(x) \tan(x) dx$

  13. $\int_0^{\pi /2} \cos ^2(x) \sin ^2(x) dx$

  14. $\int_{-π/2}^{\pi /2} \cos ^2(x) \sin ^4(x) dx$

  15. $\int _0^{\pi /4} \sec ^2(x) \tan ^2(x) dx$

  16. $\int _0 ^\pi \sin ^6(x) dx$

  17. Write a description of how to attack systematically an integation problem involving a product of powers of sines and cosines.

  18. Write a description of how to attack systematically an integation problem involving a product of powers of secants and tangents.


    Find the following integrals:

  19. $\int \sin(4x) \cos(3x) dx$  

  20. $\int \sin(4x) \sin(3x) dx$

  21. $\int \cos(4x) \cos(3x) dx$

  22. $\int \sin(5x) \cos(3x) dx$