Question

مطلوب است محاسبه انتگرال:

$$ \int _0^ \frac{ \pi }{4} cosx \sqrt{ln( \frac{ csc^{2}x}{2} )}dx= \frac{ \sqrt{ \pi } }{2} $$

Answer 1

$$ \int _0^ \frac{ \pi }{4} cosx \sqrt{ln( \frac{1}{1-cos2x} )}dx= \int _0^ \frac{ \pi }{4} cosx \sqrt{-ln(1-cos2x)} dx \wedge t=-ln(1-cos2x) \Longrightarrow dx= \frac{- e^{-t} dt}{2 \sqrt{2 e^{-t} - e^{-2t} } } \wedge 2 sin^{2} x= e^{-t} \Longrightarrow sin x= \frac{ e^{ \frac{-t}{2} } }{ \sqrt{2} } \Longrightarrow ?= \int _ { \infty } ^0 \frac{ \sqrt{2- e^{-t} } }{ \sqrt{t} } \sqrt{t} . \frac{- e^{-t} dt}{2 \sqrt{ e^{-t} (2- e^{-t} )} } = \frac{1}{2 \sqrt{2} } \int _0^ \infty \sqrt{t} . e^{ \frac{-t}{2} } dt \wedge u= \frac{t}{2} \Longrightarrow ?= \frac{1}{2 \sqrt{2} } \int _0^ \infty \sqrt{2u} e^{-u} .2du = \int _0^ \infty u^{ \frac{3}{2} -1} . e^{-u} du = \Gamma ( \frac{3}{2} )= \frac{ \sqrt{ \pi } }{2} $$