find-0-pi-2-Log-cosx-dx-

Question Number 74995 by vishalbhardwaj last updated on 05/Dec/19

find \int_{0}^{\frac{π}{2}} Log \cos x d x

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

l e t I = \int_{0}^{\frac{π}{2}} l n (c o s x) d x a n d J = \int_{0}^{\frac{π}{2}} l n (s i n x) d x

w e h s v e J =_{x = \frac{π}{2} - t} \int_{\frac{π}{2}}^{0} l n (c o s t) (- d t) = \int_{0}^{\frac{π}{2}} l n (c o s t) d t = I \Rightarrow

2 I = I + J = \int_{0}^{\frac{π}{2}} l n (c o s x s i n x) d x = \int_{0}^{\frac{π}{2}} l n (\frac{1}{2} s i n (2 x)) d x

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

\int_{0}^{\frac{π}{2}} l n (s i n (2 x)) d x =_{2 x = t} \frac{1}{2} \int_{0}^{π} l n (s i n t) d t

= \frac{1}{2} \int_{0}^{\frac{π}{2}} l n (s i n t) d t + \frac{1}{2} \int_{\frac{π}{2}}^{π} l n (s i n t) d t

\int_{\frac{π}{2}}^{π} l n (s i n t) d t =_{x = \frac{π}{2} + u} \int_{0}^{\frac{π}{2}} l n (c o s u) d u \Rightarrow

2 I = - \frac{π}{2} l n (2) + I \Rightarrow I = - \frac{π}{2} l n (2) w e h a v e p r o v e d t h a t

\int_{0}^{\frac{π}{2}} l n (c o s x) d x = \int_{0}^{\frac{π}{2}} l n (s i n x) d x = - \frac{π}{2} l n (2) .

Answered by mind is power last updated on 05/Dec/19

\int_{0}^{\frac{π}{2}} \log (\cos (x)) dx = \int_{0}^{\frac{π}{2}} \log (\sin (x)) dx, ∴ x \to \frac{π}{2} - x ∴

\int_{0}^{\frac{π}{2}} \log (\sin (2 x)) dx = \int_{0}^{π} \log (2 \sin (x) \cos (x)) = π \log (2) + \int_{0}^{\frac{π}{2}} \log (\sin (x)) dx + \int_{0}^{\frac{π}{2}} \log (\cos (x)) dx

= π \log (2) + 2 \int_{0}^{\frac{π}{2}} \log (\sin (x)) dx

u = 2 x \Rightarrow= \int_{0}^{π} \frac{\log (\sin (u)) du}{2}

\int_{0}^{π} \log (\sin (u)) du = \int_{0}^{\frac{π}{2}} \log (\sin (u)) du + \int_{\frac{π}{2}}^{π} \log (\sin (u)) du

\int_{\frac{π}{2}}^{π} \log (\sin (u)) du = \int_{0}^{\frac{π}{2}} \log (\sin (\frac{π}{2} + u)) d (u + \frac{π}{2}) = \int_{0}^{\frac{π}{2}} \log (\cos (u)) du

\Rightarrow \int_{0}^{π} \frac{\log (\sin (u)) du}{2} = \int_{0}^{\frac{π}{2}} \log (\sin (u)) du

\Leftrightarrow π \log (2) + 2 \int_{0}^{\frac{π}{2}} \log (\sin (u)) du = \int_{0}^{\frac{π}{2}} \log (\sin (u)) du

\Rightarrow \int_{0}^{\frac{π}{2}} \log (\sin (u)) du = - π \log (2)

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

e r r o r o f c a l c u l u s s i r m i n d . .