integrate ln (cosx+sinx)dx


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Question Number 46667 by Necxx last updated on 29/Oct/18

$i n t e g r a t e \ln (c o s x + s i n x) d x$
Commented by maxmathsup by imad last updated on 29/Oct/18

$l e t f (x) = \int_{0}^{x} l n (c o s t + s i n t) d t w e h a v e$ $f (x) = \int_{0}^{x} l n (\sqrt{2} c o s (t - \frac{π}{4})) d t = \frac{x}{2} l n (2) + \int_{0}^{x} l n (c o s (t - \frac{π}{4}) d t b u t$ $\int_{0}^{x} l n (c o s (t - \frac{π}{4})) d t =_{t - \frac{π}{4} = - u} \int_{\frac{π}{4}}^{\frac{π}{4} - x} l n (c o s u) d u b y p a r t s$ $= {[u l n (c o s u)]}_{\frac{π}{4}}^{\frac{π}{4} - x} - \int_{\frac{π}{4}}^{\frac{π}{4} - x} u \frac{- s i n u}{c o s u} d u$ $= \int_{\frac{π}{4}}^{\frac{π}{4} - x} u t a n (u) (d u c h a n g e m e n t t a n u = t)$ $= \int_{1}^{t a n (\frac{π}{4} - x)} t a r c t a n t \frac{d t}{1 + t^{2}} = \int_{1}^{t a n (\frac{π}{4} - x)} \frac{t a r c t a n t}{1 + t^{2}} d t$ $= {[\frac{1}{2} l n (1 + t^{2}) a r c t a n t]}_{1}^{t a n (\frac{π}{4} - x)} - \int_{1}^{t a n (\frac{π}{4} - x)} \frac{1}{2} l n (1 + t^{2}) \frac{d t}{1 + t^{2}}$ $= \frac{1}{2} (\frac{π}{4} - x) l n (1 + {t a n}^{2} (\frac{π}{4} - x)) - \frac{π}{8} l n (2) - \frac{1}{2} \int_{1}^{t a n (\frac{π}{4} - x)} \frac{l n (1 + t^{2})}{1 + t^{2}} d t$ $. . . b e c o n t i n u e d . . .$
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