0-pi-2-cos-2x-ln-sin-x-dx-pi-4-solution-1-0-pi-2-2cos-2-x-1-ln-sin-x-dx-2-0-pi-2-cos-2-x-ln-sin-x-dx-0-pi-2-

Question Number 153555 by mnjuly1970 last updated on 08/Sep/21

Ω := \int_{0}^{\frac{π}{2}} c o s (2 x) . l n (s i n (x)) d x \overset{?}{=} - \frac{π}{4}

s o l u t i o n (1)

Ω := \int_{0}^{\frac{π}{2}} (2 {c o s}^{2} (x) - 1) l n (s i n (x)) d x

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

w e k n o w t h a t : \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x \underset{e a r l i e r}{\overset{d e r i v e d}{=}} \frac{- π}{2} l n (2)

\int_{0}^{\frac{π}{2}} {c o s}^{2} (x) . l n (s i n (x)) d x \underset{p o s t s}{\overset{p r e v i o u s}{=}} - \frac{π}{4} l n (2) - \frac{π}{8}

∴ Ω := - \frac{π}{2} l n (2) - \frac{π}{4} + \frac{π}{2} l n (2)

◂ Ω = - \frac{π}{4} ▸ m . n

Answered by Ar Brandon last updated on 08/Sep/21

Ω = \int_{0}^{\frac{π}{2}} \cos (2 x) \ln (\sin x) d x

{\begin{cases} u (x) = \ln (\sin x) \\ v^{'} (x) = \cos (2 x) \end{cases} \Rightarrow {\begin{cases} u^{'} (x) = \frac{\cos x}{\sin x} \\ v (x) = \frac{\sin 2 x}{2} \end{cases}

Ω = {[\frac{\ln (\sin x) \sin 2 x}{2} - \int \cos^{2} x d x]}_{0}^{\frac{π}{2}}

= - \int_{0}^{\frac{π}{2}} \cos^{2} x d x = - \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + \cos 2 x) d x

= - \frac{1}{2} {[x + \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} = - \frac{π}{4}

Commented by mnjuly1970 last updated on 08/Sep/21

v e r y n i c e s i r b r a n d o a s a l w a y s . .

Commented by Ar Brandon last updated on 08/Sep/21

Thanks