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 Check out our web site! http://mathforum.org/dr.math
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