calculateI = ∫_0 ^(π/2) ln(cosx +sinx)dx and J =∫


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Question Number 96495 by abdomathmax last updated on 01/Jun/20

$calculateI = \int_{0}^{\frac{π}{2}} \ln (cosx + sinx) dx$ $and J = \int_{0}^{\frac{π}{2}} \ln (cosx - sinx) dx$
	Answered by Sourav mridha last updated on 02/Jun/20

	$J - I = \int_{0}^{\frac{π}{2}} l n [t a n (\frac{π}{4} - x)] d x . . . (i)$ $a n d a l s o J - I = \int_{0}^{\frac{π}{2}} l n [c o t (\frac{π}{4} - x)] d x$ $. . . . . . . . . (i i)$ $(i) + (i i) w e g e t . . J - I = 0, J = I . . . . (g)$ $n o w J + I = \int_{0}^{\frac{π}{2}} l n (c o s 2 x) d x . . . (v)$ $n o w 2 x = t, t h e n$ $2 J = \frac{1}{2} \int_{0}^{π} l n (c o s t) d t w e c a n a l s o$ $w r i t e t h i s a s - -$ $2 J = \int_{0}^{\frac{π}{2}} l n (c o s t) d t . . . . (a)$ $s o, 2 J = \int_{0}^{\frac{π}{2}} l n (s i n t) d t . . . . (b)$ $(a) + (b) 4 J = \frac{π}{2} l n (\frac{1}{2}) + \int_{0}^{\frac{π}{2}} l n (s i n (2 t)) d t$ $b y a p p l y : the s a m e a p p r o c h l i k e$ ${e q}^{n} (v) w e g e t, \int_{0}^{\frac{π}{2}} l n (s i n 2 t) d t = 2 J$ $so, 4 J = \frac{π}{2} l n (\frac{1}{2}) + 2 J$ $so, 2 J = \frac{π}{2} \ln (\frac{1}{2})$ $now f r o m (g) I = J = \frac{π}{4} l n (\frac{1}{2})$
	Commented by abdomathmax last updated on 02/Jun/20

	$thanks sir .$
	Commented by Sourav mridha last updated on 02/Jun/20
	welcome....
	Answered by mathmax by abdo last updated on 02/Jun/20

	$let f (a) = \int_{0}^{\frac{π}{2}} \ln (cosx + asinx) dx we have f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{sinx}{cosx + asinx} dx$ $= \int_{0}^{\frac{π}{2}} \frac{dx}{a + \frac{1}{tanx}} =_{tanx = u} \int_{0}^{\infty} \frac{du}{(1 + u^{2}) (a + \frac{1}{u})} = \int_{0}^{\infty} \frac{udu}{(au + 1) (u^{2} + 1)}$ $let decompose F (u) = \frac{u}{(au + 1) (u^{2} + 1)} \Rightarrow F (u) = \frac{α}{au + 1} + \frac{β u + λ}{u^{2} + 1}$ $α = \frac{- 1}{a (\frac{1}{a^{2}} + 1)} = \frac{- a^{2}}{a (1 + a^{2})} = \frac{- a}{1 + a^{2}}$ $\lim_{u \to + \infty} uF (u) = 0 = \frac{α}{a} + β \Rightarrow β = - \frac{α}{a} = \frac{1}{1 + a^{2}}$ $F (0) = 0 = α + λ \Rightarrow λ = \frac{a}{1 + a^{2}} \Rightarrow F (u) = - \frac{a}{(1 + a^{2}) (au + 1)} + \frac{1}{1 + a^{2}} \times \frac{u + a}{u^{2} + 1}$ $\Rightarrow \int_{0}^{\infty} F (u) du = - \frac{a}{1 + a^{2}} \int_{0}^{\infty} \frac{du}{au + 1} + \frac{1}{2 (1 + a^{2})} \int_{0}^{\infty} \frac{2 u}{u^{2} + 1} + \frac{a}{1 + a^{2}} \int_{0}^{\infty} \frac{du}{u^{2} + 1}$ $= \frac{1}{1 + a^{2}} {[\ln \sqrt{u^{2} + 1} - \ln (au + 1)]}_{0}^{\infty} + \frac{π a}{2 (1 + a^{2})}$ $= \frac{1}{1 + a^{2}} {[\ln (\frac{\sqrt{1 + u^{2}}}{au + 1})]}_{0}^{\infty} + \frac{π a}{2 (1 + a^{2})} = - \frac{\ln (a)}{1 + a^{2}} + \frac{π a}{2 (1 + a^{2})} \Rightarrow$ $f (a) = - \int_{0}^{a} \frac{lnt}{1 + t^{2}} dt + \frac{π}{4} \ln (1 + a^{2}) + c but$ $f (0) = c = - \frac{π}{2} \ln (2) \Rightarrow f (a) = \frac{π}{4} \ln (1 + a^{2}) - \frac{π}{2} \ln (2) - \int_{0}^{a} \frac{lnt}{1 + t^{2}} dt$ $\int_{0}^{\frac{π}{2}} \ln (cosx + sinx) dx = f (1) = \frac{π}{4} \ln (2) - \frac{π}{2} \ln (2) - \int_{0}^{1} \frac{lnt}{1 + t^{2}} dt (\to t = \tan θ)$ $= - \frac{π}{4} \ln (2) - \int_{0}^{\frac{π}{2}} \ln (\tan θ) d θ = - \frac{π}{4} \ln (2) - 0 (because \int_{0}^{\frac{π}{2}} \ln (\cos θ) d θ = \int_{0}^{\frac{π}{2}} \ln (\sin θ) d θ) \Rightarrow$ $\Rightarrow \int_{0}^{\frac{π}{2}} \ln (cosx - sinx) dx = - \frac{π}{4} \ln (2)$
	Commented by mathmax by abdo last updated on 02/Jun/20

	$error of typo \int_{0}^{\frac{π}{2}} \ln (cosx + sinx) dx = - \frac{π}{4} \ln 2$
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