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

Apr 23, 2018

$\pi \ln | \sec 1 + \tan 1 |$

#### Explanation:

We have,
'
${\int}_{0}^{1} 2 \pi \frac{x}{\cos} x \mathrm{dx}$

$= 2 \pi \cdot {\int}_{0}^{1} \frac{x}{\cos} x \mathrm{dx}$ [Constant Part is Taken out]

$= 2 \pi \cdot {\int}_{0}^{1} x \sec x \mathrm{dx}$

$= 2 \pi \cdot {\left[\frac{1}{2} {x}^{2} \ln | \sec x + \tan x |\right]}_{0}^{1}$ [Because $\int \sec x \mathrm{dx} = \ln | \sec x + \tan x | + C$]

$= 2 \pi \cdot \left\{\frac{1}{2} {1}^{2} \ln | \sec 1 + \tan 1 | - \frac{1}{2} {0}^{2} \ln | \sec 0 + \tan 0 |\right\}$

$= 2 \pi \cdot \left(\frac{1}{2} \ln | \sec 1 + \tan 1 |\right)$

$= \pi \ln | \sec 1 + \tan 1 |$

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:
$\int \left(\frac{x}{\cos} \left(x\right)\right) = i \left(L {i}_{2} \left(- i {e}^{i x}\right) - L {i}_{2} \left(i {e}^{i x}\right)\right) + x \left(\ln \left(1 - i {e}^{i x}\right) - \ln \left(1 + {e}^{i x}\right)\right)$
(Compair: https://www.wolframalpha.com/input/?i=int+x%2Fcos(x))

${\int}_{0}^{1} \left(2 \pi \frac{x}{\cos} \left(x\right) \mathrm{dx}\right) \approx 4.24$