calculate ∫_0 ^(π/2) ln(cosx+sinx)ex


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 47112 by maxmathsup by imad last updated on 04/Nov/18

$c a l c u l a t e \int_{0}^{\frac{π}{2}} l n (c o s x + s i n x) e x$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

$0 < \int_{0}^{\frac{π}{2}} l n (s i n x + c o s x) < \frac{π}{2} \times 0.5$ $0 < I < \frac{π}{4}$
Commented by prof Abdo imad last updated on 06/Nov/18

$l e t I = \int_{0}^{\frac{π}{2}} l n (c o s x + s i n x) d x \Rightarrow$ $I = \int_{0}^{\frac{π}{2}} l n (\sqrt{2} c o s (x - \frac{π}{4})) d x = \frac{π}{4} l n (2) + \int_{0}^{\frac{π}{2}} l n (c o s (x - \frac{π}{4})) d x$ $b u t \int_{0}^{\frac{π}{2}} l n (c o s (x - \frac{π}{4})) d x =_{\frac{π}{4} - x = t} - \int_{\frac{π}{4}}^{- \frac{π}{4}} l n (c o s t) d t$ $= \int_{- \frac{π}{4}}^{\frac{π}{4}} l n (c o s t) d t = 2 \int_{0}^{\frac{π}{4}} l n (c o s t) d t$ $l e t A = \int_{0}^{\frac{π}{4}} l n (c o s t) d t a n d B = \int_{0}^{\frac{π}{4}} l n (s i n t) d t$ $A + B = \int_{0}^{\frac{π}{4}} l n (c o s t s i n t) d t = \int_{0}^{\frac{π}{4}} l n (\frac{s i n (2 t)}{2}) d t$ $= - \frac{π}{4} l n (2) + \int_{0}^{\frac{π}{4}} l n (s i n (2 t)) d t$ $=_{2 t = u} - \frac{π}{4} l n (2) + \frac{1}{2} \int_{0}^{\frac{π}{2}} l n (s i n u) d u$ $= - \frac{π}{4} l n (2) - \frac{π}{4} l n (2) = - \frac{π}{2} l n (2) a l s o$ $A - B = \int_{0}^{\frac{π}{4}} l n (c o s t) d t - \int_{0}^{\frac{π}{4}} l n (s i n t) d t b u t$ $\int_{0}^{\frac{π}{4}} l n (s i n t) d t =_{t = \frac{π}{2} - x} - \int_{\frac{π}{2}}^{\frac{π}{4}} l n (c o s x) d x$ $= \int_{\frac{π}{4}}^{\frac{π}{2}} l n (c o s x) d x = \int_{0}^{\frac{π}{2}} l n (c o s x) d x - \int_{0}^{\frac{π}{4}} l n (c o s x) d x$ $= - \frac{π}{2} l n (2) - A \Rightarrow A - B = - \frac{π}{2} l n 2 - A \Rightarrow$ $2 A = B - \frac{π}{2} l n (2) \Rightarrow 2 A = - \frac{π}{2} l n (2) - A - \frac{π}{2} l n (2)$ $\Rightarrow 2 A = - A - π l n (2) \Rightarrow 3 A = - π l n (2) \Rightarrow$ $A = - \frac{π}{3} l n (2) \Rightarrow I = - \frac{2 π}{3} l n (2) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18
	$ln(cosx+sinx) ∫_0 ^(π/2) ln{(√2) ×((1/(√2))sinx+(1/(√2))cosx)dx (1/2)ln2∫_0 ^(π/2) dx+∫_0 ^(π/2) ln{sin((π/4)+x)}dx I_1 =((1/2)ln2)×(π/2)=(π/4)ln2 I_2 wait...$
	$l n (c o s x + s i n x)$ $\int_{0}^{\frac{π}{2}} l n {\sqrt{2} \times (\frac{1}{\sqrt{2}} s i n x + \frac{1}{\sqrt{2}} c o s x) d x$ $\frac{1}{2} l n 2 \int_{0}^{\frac{π}{2}} d x + \int_{0}^{\frac{π}{2}} l n {s i n (\frac{π}{4} + x)} d x$ $I_{1} = (\frac{1}{2} l n 2) \times \frac{π}{2} = \frac{π}{4} l n 2$ $I_{2}$ $w a i t . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum