find-I-0-pi-4-ln-cosx-dx-and-J-0-pi-4-ln-sinx-dx-

Question Number 151246 by mathmax by abdo last updated on 19/Aug/21

find I = \int_{0}^{\frac{π}{4}} \ln (cosx) dx and J = \int_{0}^{\frac{π}{4}} \ln (sinx) dx

Answered by qaz last updated on 19/Aug/21

\int_{0}^{π / 4} lnsin xdx

= \frac{1}{2} \int_{0}^{π / 2} lnsin \frac{x}{2} dx

= \frac{1}{4} \int_{0}^{π / 2} \ln \frac{1 - \cos x}{2} dx

= \frac{1}{4} \int_{0}^{π / 2} \ln (\sin xtan \frac{x}{2}) dx - \frac{1}{4} \int_{0}^{π / 2} \ln 2 dx

= - \frac{π}{8} \ln 2 + \frac{1}{4} \int_{0}^{π / 2} lntan \frac{x}{2} dx - \frac{π}{8} \ln 2

= - \frac{π}{4} \ln 2 + \frac{1}{2} \int_{0}^{π / 4} lntan xdx

= - \frac{π}{4} \ln 2 + \frac{1}{2} \int_{0}^{1} \frac{lnx}{1 + x^{2}} dx

= - \frac{π}{4} \ln 2 + \frac{1}{2} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{2 n} lnxdx

= - \frac{π}{4} \ln 2 + \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{{(2 n + 1)}^{2}}

= - \frac{π}{4} \ln 2 - \frac{1}{2} G

- - - - - - - - - - - - - - -

\int_{0}^{π / 4} lncos xdx + \int_{0}^{π / 4} lnsin xdx

= \int_{0}^{π / 4} \ln (\frac{1}{2} \sin 2 x) dx

= \int_{0}^{π / 4} \ln \frac{1}{2} dx + \frac{1}{2} \int_{0}^{π / 2} lnsin xdx

= - \frac{π}{2} \ln 2

\Rightarrow \int_{0}^{π / 4} lncos xdx = - \frac{π}{2} \ln 2 - (- \frac{π}{4} \ln 2 - \frac{1}{2} G) = \frac{1}{2} G - \frac{π}{4} \ln 2

Commented by peter frank last updated on 19/Aug/21

thank you