Techniques of Integration - Change of VariablesDate: 02/17/99 at 17:46:41 From: jenny Subject: Techniques of integration Hello. I would like to know how to solve this question. (integral sign) sin 2x/ sq rt (9-cos^4 x) dx How should I begin? Jen Date: 02/18/99 at 00:53:14 From: Doctor Luis Subject: Re: Techniques of integration You can approach this question more easily if you use the identity sin(2x) = 2sin(x)cos(x), and if you rewrite your integral in a more suggestive way, like this: / | 2 sin(x) cos(x) | ----------------------- dx | sqrt(9 - (cos^2(x))^2) / At this point it is clear that a change of variables will do the trick. Let cos^2(x) = 3u. Then, differentiating implicitly, 2cos(x)*(-sin(x))dx = 3du This means that 2cos(x)sin(x) = -3du, and so, rewriting our integral in terms of u, we have / / | -3 du | du | ---------------- = - | ------------- | sqrt(9 - 9 u^2) | sqrt(1-u^2) / / But this last integral is already known. It's nothing more than the inverse cosine function. So, / | du - | ------------- = arccos(u) = arccos((1/3)cos^2(x)) | sqrt(1-u^2) / Notice that we have expressed our final answer in terms of the original variable of integration, using the fact that u = (1/3)cos^2(x) . Obviously, I have neglected the arbitrary constant of integration, but you can add that at any time. Now, to show that the integral of -1/sqrt(1-u^2) is the arccos(u) function, what you can do is check that the derivative of the arccos(u) function is -1/sqrt(1-u^2). We can do that as follows: Let y = arccos(x); then x = cos(y). By implicit differentiation on this last equation you can obtain, 1 = -sin(y) * dy/dx (notice we used the chain rule here) Solving for dy/dx, you get -1 dy/dx = --------- sin(y) Expressing sin(y) in terms of cos(y) (use the identity sin^2(y) + cos^2(y) = 1 for this step) we get: -1 dy/dx = ----------------- sqrt(1-cos^2(y)) But, by definition, x = cos(y). Therefore, -1 dy/dx = ------------- sqrt(1-x^2) Now, y was just y = arccos(x). Therefore, we have proven that the derivative of the arccos(x) function is -1/sqrt(1-x^2) . Feel free to ask again if you have any other questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/