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

2 Answers
Apr 23, 2018

#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.

Apr 30, 2018

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#