Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Integration


Date: 02/17/99 at 17:48:47
From: crys
Subject: Integration

Can you help me with

  (integral sign) x tan^2 x dx   ?

I think you have to use some kind of substitution or maybe trigonometry 
identities 

Thank you. 


Date: 02/18/99 at 01:28:50
From: Doctor Luis
Subject: Re: Integration

The first thing that comes to mind is integration by parts. My primary 
motivation for this method is the factor of x that appears on the 
integrand, because I know that differentiating that factor will give 
me a constant.

Quick review: integration by parts is essentially based on the 
following formula

    /                   /
   |                   |
   |  u dv  =  u*v  -  | v du
   |                   |
  /                   /

In this case the substitutions should be

     u = x              ----->     du = dx

    dv = tan^2(x) dx    ----->      v = ?

Note that there is a small problem here. We need to integrate the 
function tan^2(x) in order to obtain v from dv. But this does not 
really present that much of a problem, if you remember that 
tan(x) = sin(x)/cos(x) and that cos^2(x)+sin^2(x) = 1.

       /                /
      |  sin^2(x)      |  1 - cos^2(x)
 v =  | --------- dx = | ------------- dx
      |  cos^2(x)      |    cos^2(x)
     /                /

        /
       |
    =  | (sec^2(x)-1)dx = tan(x)-x
       |
      /

Therefore, v = tan(x)-x. Now, substituting into our "integration by 
parts formula," you get

    /                                     /
   |                                     |
   | (x)*(tan^2(x)dx) = (x)*(tan(x)-x) - | (tan(x)-x)*(dx)
   |                                     |
  /                                     /

where I have enclosed the substitutions we made in parentheses.

Now, the second term on the integral on the righthand side of the
equation presents no problem. It's just (1/2)x^2, but the first term
seems a bit more difficult. However, if you rewrite tan(x) as
sin(x)cos(x)/cos^2(x), you can see that we almost have a complete
differential; the derivative of cos^2(x) is 2*cos(x)*(-sin(x)).
So, concentrating on the integral of tan(x), we have:

    /            /
   |            |  sin(x)cos(x)
   | tan(x)dx = | ------------- dx
   |            |   cos^2(x)
  /            /

If you make the substitution u = cos^2(x), which implies that
du = 2cos(x)*(-sin(x))dx or cos(x)sin(x)dx = -du/2), then this last
integral becomes

    /                        /
   |  sin(x)cos(x)       1  |  du
   | ------------- =  - --- | ----- = -(1/2)ln(u) = -(1/2)ln(cos^2(x))
   |    cos^2(x)         2  |   u  
  /                        /

Therefore the integral of tan(x) is just -(1/2)ln(cos^2(x)) and so our 
original integral becomes:

    /                                     /
   |                                     |
   | (x)*(tan^2(x)dx) = (x)*(tan(x)-x) - | (tan(x)-x)*(dx)
   |                                     |
  /                                     /

                      = x*tan(x)-x^2 - (-(1/2)ln(cos^2(x))-(1/2)x^2)

                      = x*tan(x) - (1/2)x^2 + (1/2)ln(cos^2(x))

I hope this explanation helps, although it would have been better if 
you had included an explanation of how you attempted this problem and 
where you got stuck. (I did not know how or why you got stuck, so I 
tried to explain everything.) Feel free to reply if you have any 
questions about this answer.

Sincerely,

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Trigonometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/