Integration of Trigonometric Function by Substitution

Date: 03/18/98 at 20:05:21
From: Jennifer Haas
Subject: derivatives

Int[tan 2x dx]

I get confused when I need to do a u-substitution, or if this problem 
actually needs one. At what point of the problem do I actually take 
the derivative? Can you help me out please?

Thank you very much.

Jennifer

Date: 03/19/98 at 12:26:51
From: Doctor Sam
Subject: Re: derivatives

Jennifer,

Substitutions can be confusing! Lots and lots of practice will help. 
The other help is to memorize the standard formulas for 
antideritivatives so that you will more easily recognize them when you 
see them.

And substitutions aren't magic. It is often hard to tell what the best 
substitution might be. I often find myself trying one substitution, 
working the problem for a while, and then deciding that I am not 
getting anywhere so that I need to try a different substitution.

This problem is a good example of that. It is often useful to make the 
substitution u = 2x in a problem like this so that you have sin(u) or 
tan(u). . . a trig function of a single variable rather than a trig 
function of another function.

But in this case, if you try u = 2x and du = 2dx, you will transform 
the integral this way:  

     INT(tan(2x)dx) = (1/2)INT(tan(u)du)

and this isn't helpful (to me) because I don't know the antiderivative 
of tan(u).

So I would go back to the beginning. There really aren't any other 
possible substitutions in this problem so I would look to 
trigonometry. Is there any other way to write tan(2x)?  

Yes. The tangent is the ratio of the sine and the cosine. So I can 
rewrite the problem as:
                            sin(2x)
     INT(tan(2x)dx)  =  INT(-------  dx)
                            cos(2x)

And I know that sines and cosines are derivatives of each other. That 
suggests (to me) letting u = cos(2x). Differentiate to get

     du = -2sin(2x)dx,

so that 

     sin(2x)dx  = -(1/2)du.
     
Now substitute:

         sin(2x)                    du
     INT(------- dx)  = -(1/2) INT(----)
         cos(2x)                    u

If you have memorized your standard antiderivatives, you will 
recognize this integral as a natural logarithm. So:

                  du
     -(1/2) INT (----) = -(1/2) ln |u| + c,
                  u

Substituting cos(2x) back in for u, I find that

     INT(tan(2x)dx) = -(1/2) ln |cos(2x)| + c.

I hope this helps!

- Doctor Sam

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