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### Integration of Trigonometric Function by Substitution

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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
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```
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

Check out our web site! http://mathforum.org/dr.math
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Associated Topics:
High School Calculus

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