integrate-ln-cosx-sinx-dx-

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

\dots b e c o n t i n u e d \dots