Rather tricky integral. I liked it :)

freemind

Rather tricky integral. I liked it :)

**Posted:** Sun Mar 02, 2008 5:12 pm

_________________
The fate of equilibrium is to end the eternity...

kunny

Let $x = \frac {\pi}{4} - \theta \Longrightarrow dx = - \theta ,\ 2x = \frac {\pi}{2} - 2\theta$ ,

$I = \int_{\frac {\pi}{4}}^0 (\cos ^ 6 2\theta + \sin ^ 6 2\theta)\ln \left [1 + \tan \left(\frac {\pi}{4} - \theta \right)\ri...$

$= \int_0^{\frac {\pi}{4}} (\cos ^ 6 2\theta + \sin ^ 6 2\theta)\ln \frac {2}{1 + \tan \theta} d\theta$

$= (\ln 2)\int_0^{\frac {\pi}{4}} (\cos ^ 6 2\theta + \sin ^ 6 2\theta)\ d\theta - I$

$2I = (\ln 2)\int_0^{\frac {\pi}{4}} (\cos ^ 6 2\theta + \sin ^ 6 2\theta)\ d\theta$

Here, we have :

$\sin ^ 6 \theta + \cos ^ 6 \theta = (\sin ^ 2 \theta + \cos ^ 2 \theta)^3 - 3\sin ^ 2 \theta \cos ^ 2 \theta (\sin ^ 2 \theta...$
$= 1 - 3\sin ^ 2 \theta \cos ^ 2 \theta$
$= 1 - \frac {3}{4}\sin ^ 2 2\theta$
$= 1 - \frac {3}{4}\cdot \frac {1 - \cos 4\theta}{2}$
$= \frac {1}{8}(5 + 3\cos 4\theta)$

$\therefore 2I = \frac {\ln 2}{8}\int_0^{\frac {\pi}{4}} (5 + 3\cos 8\theta)\ d\theta = \frac {5}{32}\pi \ln 2$ , yielding $I = \frac {5}{64}\pi \ln2$ .

**Posted:** Sun Mar 02, 2008 5:50 pm

_________________
Today's calculation of Integral Digest
Hang in there, students.