Question

What is `int_(0)^(1) 2(pi)x/(cosx)dx`?

Answer 1

`piln|sec 1 + tan 1|`

Explanation:

We have,
'
`int_0^1 2pix/cos x dx`

`= 2 pi * int_0^1 x/cosx dx` [Constant Part is Taken out]

`= 2 pi * int_0^1 x sec x dx`

`= 2 pi * [1/2x^2ln|sec x + tan x|]_0^1` [Because `int sec x dx = ln|sec x + tanx | +C`]

`= 2 pi * {1/2 1^2 ln|sec 1 + tan 1| - 1/2 0^2ln|sec0 + tan 0|}`

`= 2pi * (1/2ln|sec 1 + tan 1|)`

`= piln|sec 1 + tan 1|`

Hope this helps.

Answer 2

Are you sure that there isn't a typo in the question?

Explanation:

...because the answer looks like this:
`int(x/cos(x))=i(Li_2(-ie^(ix))-Li_2(ie^(ix)))+x(ln(1-ie^(ix))-ln(1+e^(ix)))`
(Compair: https://www.wolframalpha.com/input/?i=int+x%2Fcos(x))

Therefore, the over-all answer is:
`int_0^1(2pix/cos(x)dx)~~4.24`