# Definite Integration (1 Viewer)

#### juantheron

##### Active Member
$\bg_white Evaluation of \int^{\frac{\pi}{2}}_{0}\cos x\cdot \ln(\cos x)dx$

#### blyatman

##### Well-Known Member
If you integrate by parts, but take the limit as x->pi/2 from below, you will get the correct result: ln(2)-1.

Btw, how do you type those LaTeX-like equations in your reply? I can draw up the working out if you'd like.

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#### fan96

##### 617 pages
You can use LaTeX with the tags [Xtex] [X/tex] (remove the red X's).

for example, [Xtex] \left( \frac{\pi^2}{6} \right) [X/tex] gives:

$\bg_white \left( \frac{\pi^2}{6} \right)$

#### blyatman

##### Well-Known Member
You can use LaTeX with the tags [Xtex] [X/tex] (remove the red X's).

for example, [Xtex] \left( \frac{\pi^2}{6} \right) [X/tex] gives:

$\bg_white \left( \frac{\pi^2}{6} \right)$
Thanks!

#### blyatman

##### Well-Known Member
Here you go:

$\bg_white \int_0^{\frac{\pi}{2}}\cos x \log(\cos x)\,dx\\= \lim_{\alpha \to \left(\frac{\pi}{2}\right)^{-}}\int_0^{\alpha}\cos x \log(\cos x)\,dx \\ = \lim_{\alpha \to \left(\frac{\pi}{2}\right)^{-}}\left(\left[\sin x \log(\cos x)\right]_0^{\alpha}+\int_0^{\alpha}\frac{\sin^2 x}{\cos x}\,dx\right)\\=\lim_{\alpha \to \left(\frac{\pi}{2}\right)^{-}}\left(\log(\cos \alpha) + \left[\log\left|\tan x + \sec x\right|-\sin x\right]_0^{\alpha}\right)\\= \lim_{\alpha \to \left(\frac{\pi}{2}\right)^{-}}\left(\log(\cos \alpha) + \log(\tan \alpha + \sec \alpha)-1\right)\\= \lim_{\alpha \to \left(\frac{\pi}{2}\right)^{-}}\left[\log(\sin \alpha +1)-1\right]\\= \log(2)-1$

Line 4: See https://socratic.org/questions/how-do-you-find-the-integral-of-int-sin-x-tan-x-dx

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