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