calculate-0-pi-4-ln-cosx-dx-and-0-pi-4-ln-sinx-dx-

Question Number 70262 by mathmax by abdo last updated on 02/Oct/19

c a l c u l a t e \int_{0}^{\frac{π}{4}} l n (c o s x) d x a n d \int_{0}^{\frac{π}{4}} l n (s i n x) d x

Commented by mathmax by abdo last updated on 06/Oct/19

l e t I = \int_{0}^{\frac{π}{4}} l n (c o s x) d x a n d J = \int_{0}^{\frac{π}{4}} l n (s i n x) d x

I + J = \int_{0}^{\frac{π}{4}} l n (c o s x + s i n x) d x = \int_{0}^{\frac{π}{4}} l n (\sqrt{2} c o s (x - \frac{π}{4})) d x

= \frac{π}{8} l n (2) + \int_{0}^{\frac{π}{4}} l n (c o s (\frac{π}{4} - x)) d x b u t

\int_{0}^{\frac{π}{4}} l n (c o s (\frac{π}{4} - x)) d x =_{\frac{π}{4} - x = t} \int_{0}^{\frac{π}{4}} l n (c o s t) d t = I \Rightarrow

I + J = \frac{π}{8} l n (2) + I \Rightarrow J = \frac{π}{8} l n (2)

I = \int_{0}^{\frac{π}{4}} l n (c o s x) d x =_{x = \frac{t}{2}} \frac{1}{2} \int_{0}^{\frac{π}{2}} l n (c o s (\frac{t}{2})) d t \Rightarrow

2 I = \int_{0}^{\frac{π}{2}} l n (s i n (\frac{π}{2} - \frac{t}{2})) d t =_{\frac{π - t}{2} = u} \int_{\frac{π}{2}}^{\frac{π}{4}} l n (s i n u) (- 2) d u

= 2 \int_{\frac{π}{4}}^{\frac{π}{2}} l n (s i n u) d u = 2 {\int_{\frac{π}{4}}^{0} l n (s i n u) d u + \int_{0}^{\frac{π}{2}} l n (s i n u) d u}

= 2 {- J - \frac{π}{2} l n (2)} = - 2 J - π l n (2) = - \frac{π}{4} l n (2) - π l n (2)

= - \frac{5 π}{4} l n (2) \Rightarrow I = - \frac{5 π}{8} l n (2)