Question

∫cos(cos2x)

Answer 1

sure, like that ?

Answer 2

\[\int\cos(\cos(2x))dx\]or\[\int\cos(\cos^2x)dx\]???

Answer 3

use power series

Answer 4

I wouldn't even know where to begin @Zarkon but wolfram refuses to do either

Answer 5

writing as a power series and integrating term by term is the only way I see doing either of those problems

Answer 6

∫cos(cos2x) is right

Answer 7

@jitu that doesn't answer my question
is the 2 an exponent or is it in the argument?

and as far as the power series approach I don't see how to integrate a power series of a power series, which is what seems to be required

Answer 8

2 is argument

Answer 9

\[\cos(\cos(2x))\]

\[=1-2 \left(x-\frac{\pi }{4}\right)^2+\frac{10}{3} \left(x-\frac{\pi }{4}\right)^4-\frac{148}{45} \left(x-\frac{\pi }{4}\right)^6+\frac{914}{315} \left(x-\frac{\pi }{4}\right)^8+O\left[x-\frac{\pi }{4}\right]^9\]

Answer 10

what is O means

Answer 11

http://en.wikipedia.org/wiki/Big_O_notation

Answer 12

I cheated and used

Series[Cos[Cos[2 x]], {x, Pi/4, 8}]

from within Mathematica 8

Answer 13

give the exact solution

Answer 14

if limit is 0 to pi/6 then what is answer

Answer 15

Mathematica still can't give an exact answer (with those limits) so I'm not going to try

Answer 16

any one can give reply of this

Answer 17

Zarkon is really probably the best mathematician on this site, and mathematica fails, so this integral is pretty darn hard. If you want to try a more scholarly site there is this one:
http://math.stackexchange.com/
You may get someone who specializes in whatever fancy analysis is required to do this integral.