Integral of Cos^2
Date: 2/6/96 at 18:26:12 From: Anonymous Subject: Anti-derivative for Calculus I can't seem to find the Anti-derivative of (cos X)^2. I have tried to find my own solution, which is: (-(cos X)^3)/(3sin x). I would like to know some more about this. Thank you!
Date: 6/13/96 at 13:58:39 From: Doctor Charles Subject: Re: Anti-derivative for Calculus Well: d/dx[(-(cos x)^3)/(3 sin x)] = (9(sin(x)cos(x))^2 -3 (cos(x))^4))/(9(sin(x)^2) = (cos(x))^2 * (1 - (1/3) (cot(x))^2) so I am afraid that your solution doesn't work. It looks as though you have used a substitution without doing it rigorously enough. Here's how it should be done. cos(2x) = (cos(x))^2 - (sin(x))^2 cos(2x) = 2 cos(x)^2 - 1 (cos(x))^2 = 1/2 * (cos(2x) + 1) if intgl() stands for the integral (anti-derivative) of a function then: intgl(cos(2x)) = 1/2 * sin(2x) + 2C so intgl((cos(x))^2) = 1/4 sin(2x) + 1/2 x + C -Doctor Charles, The Math Forum
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum