Question

how do i integrate cos^2x ?

Answer 1

standard trick is to rewrite using one of those trig reduction formulas that you thought you would never use

Answer 2

\[\cos^2(x)=\frac{1}{2}(\cos(2x)+1)\] integrate term by term

Answer 3

would the answer be 1/20 [1/2 sin 2x ] ?

Answer 4

no i don't think so

Answer 5

\[\int 1dx=x\]
\[\int \cos(2x)dx=-\frac{1}{2}\sin(2x)\]

Answer 6

sorry last one is wrong should be
\[\int \cos(2x)dx=\frac{1}{2}\sin(2x)\]

Answer 7

ohh, i took out the one outside.

Answer 8

should get as a final answer
\[\frac{x}{2}+\frac{1}{2}\sin(2x)\]

Answer 9

i gt 1/40 sin 2x + 1/20x

Answer 10

or if you like you can rewrite as
\[\frac{x}{2}+\frac{1}{2}\sin(x)\cos(x)\] by that double angle formula. either way

Answer 11

damn another mistake, hold on

Answer 12

\[\frac{x}{2}+\frac{1}{4}\sin(2x)\] is correct

Answer 13

i assume the zeros in your answer are typos.

Answer 14

so the answer would be pi/2?

Answer 15

the answer is a function, or rather a class of funcitons, not a number

Answer 16

sorry i have to substitute pi and 0, as the last part.

Answer 17

thanks a lot for your help !