- #1
iamsmooth
- 103
- 0
Homework Statement
[tex]
\int \! \cos^2xdx[/tex]
The Attempt at a Solution
Letting u = x2 and du/dx = 2x
To get 2x into the integral, I multiply the outside by 1/2x?
[tex]
\frac{1}{2x} \int \! \cos (u) du [/tex]
[tex]
\frac{1}{2x} \sin (x^2) 2x + C[/tex]
Even if I play around with this, I don't get anywhere near the correct answer, which is:
[tex] \frac{1}{2}x + \frac{1}{4}\sin (2x) + C[/tex]
There's something I'm not getting. Can someone please help? Thanks.