Question

Evaluate the integral: The integral of cos^2(x) dx.

Picture attached! If you guys could help me out, that'd be so appreciated! Thanks!

Answer 1

cosx.x-x^2/2

Answer 2

We will use the rule \[P u'v = uv - P uv'\] where P stands for primitive.
So choosing u = cos(x) and v = cos(x) we have:
\[P \cos(x) \cos(x) = sen(x)\cos(x) - P sen(x)(-sen(x))\]
\[P \cos(x) \cos(x) = sen(x)\cos(x) + P sen(x)sen(x)\]
Now you may think, well this gets us no where. But remember:
\[sen^2(x) = 1-\cos^2(x)\]

So switching that out for the sen2,
\[P \cos(x) \cos(x) = sen(x)\cos(x) + P 1-\cos^2(x)\]
\[P \cos^2(x)) = sen(x)\cos(x) + P 1- P\cos^2(x)\]
But P 1 = x! So,
\[2P \cos^2(x)) = sen(x)\cos(x) + x\]
\[P \cos^2(x)) = {sen(x)\cos(x) + x \over 2}\]
And so

\[\int\limits_{}^{}\cos^2(x) dx = {sen(x)\cos(x) + x \over 2}\]

Or if you prefer, \[\int\limits\limits_{}^{}\cos^2(x) dx = {sen(2x) \over 4} + {x \over 2}\]

Answer 3

Plus the arbitrary constant, evidently.

Answer 4

thank you! i got it now.