Question

integral of (cos theta)^2 d theta?

Answer 1

\[\int\limits \text{Cos}[\theta ]^2 \, d\theta \]

Answer 2

yes

Answer 3

could you please look at my previous question? I type the notation there

Answer 4

sorry imrammeah, it was wrong

Answer 5

trick is to write
\[\cos^2(\theta)=\frac{1}{2} \cos(2\theta)+\frac{1}{2}\]

Answer 6

could you please explain, where did you derive that from?

Answer 7

back of a text book

Answer 8

that is how i do all integration problems because they are stultifyingly boring tricks

Answer 9

\[\color{blue}{\text{myininaya will explain}}\]

Answer 10

ok do you remember
that cos(x+x)=cos(2x)=cosxcosx-sinxsinx=cos^2x-sin^2x
(but sin^2x=1-cos^2x)
cos^2x-sin^2x=cos^2x-(1-cos^2x)=2cos^2x+1
ok so we have cos(2x)=2cos^2x+1
solve this for cos^2x

Answer 11

see! you think i can remember that?

Answer 12

ok and i made a little type up there
cos2x=2cos^2x-1

Answer 13

use U substitution. \[\int\limits_{}^{} (u)^{2} du\]\[\cos \theta=u\]\[du=-\sin \theta\] \[=u^3/3\]\[=(-\sin \theta)/3\]

Answer 14

rsmith, where do I find sin theta in the problem?

Answer 15

guys, why can't I find the "good answer" button anywhere?

Answer 16

using U substitution you take the derivative of what you take out for U hence the cos theta becoming -sin theta

Answer 17

\[\int\limits_{}^{} (\frac{1}{2}\cos(2x)+\frac{1}{2}) dx=\frac{1}{4}\sin(2x)+\frac{1}{2}x+C\]

Answer 18

ok, is this the answer:\[1/2\theta + 1/4 \sin 2\theta + C\]

Answer 19

yes!

Answer 20

ok, thank you, still cant find the button

Answer 21

well then. haha i guess i was wrong :(

Answer 22

^^^^^
||||||||
the button is here

Answer 23

i'll give her one

Answer 24

btw if you are doing lots of these the back of your text is really your best friend. other than here of course

Answer 25

thanks for the tip.

Answer 26

hey aftercall, does your book have the answers in them?