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Question Number 109343 by 150505R last updated on 22/Aug/20

Commented by 150505R last updated on 22/Aug/20

$p l e a s e s o l v e$
	Answered by mathmax by abdo last updated on 23/Aug/20
	$A =∫_0 ^(1/2) ((ln(1−x))/(2x^2 −2x +1))dx ⇒A =_(x=(t/2)) ∫_0 ^1 ((ln(1−(t/2)))/(2((t^2 /4))−2((t/2))+1))(dt/2) =∫_0 ^1 ((ln(2−t))/(2{ (t^2 /2)−t +1})) dt =∫_0 ^1 ((ln(2−t))/(t^2 −2t +2)) dt =_(t=1−u) ∫_1 ^0 ((ln(2−1+u))/((1−u)^2 −2(1−u) +2))(−du) =∫_0 ^1 ((ln(1+u))/(u^2 −2u+1−2+2u +2)) du =∫_0 ^1 ((ln(1+u))/(1+u^2 )) du =_(u=tanθ) ∫_0 ^(π/4) ((ln(1+tanθ))/(1+tan^2 θ)) (1+tan^2 θ)dθ =∫_0 ^(π/4) ln(1+((sinθ)/(cosθ)))dθ =∫_0 ^(π/4) ln(cosθ +sinθ)dθ−∫_0 ^(π/4) ln(cosθ)dθ and all those integrals are known in the platform )i dont remember the values)....$
	$A = \int_{0}^{\frac{1}{2}} \frac{\ln (1 - x)}{{2 x}^{2} - 2 x + 1} dx \Rightarrow A =_{x = \frac{t}{2}} \int_{0}^{1} \frac{\ln (1 - \frac{t}{2})}{2 (\frac{t^{2}}{4}) - 2 (\frac{t}{2}) + 1} \frac{dt}{2}$ $= \int_{0}^{1} \frac{\ln (2 - t)}{2 {\frac{t^{2}}{2} - t + 1}} dt = \int_{0}^{1} \frac{\ln (2 - t)}{t^{2} - 2 t + 2} dt$ $=_{t = 1 - u} \int_{1}^{0} \frac{\ln (2 - 1 + u)}{{(1 - u)}^{2} - 2 (1 - u) + 2} (- du)$ $= \int_{0}^{1} \frac{\ln (1 + u)}{u^{2} - 2 u + 1 - 2 + 2 u + 2} du = \int_{0}^{1} \frac{\ln (1 + u)}{1 + u^{2}} du$ $=_{u = \tan θ} \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \tan θ)}{1 + \tan^{2} θ} (1 + \tan^{2} θ) d θ$ $= \int_{0}^{\frac{π}{4}} \ln (1 + \frac{\sin θ}{\cos θ}) d θ = \int_{0}^{\frac{π}{4}} \ln (\cos θ + \sin θ) d θ - \int_{0}^{\frac{π}{4}} \ln (\cos θ) d θ$ $and all those integrals are known in the platform$ $) i dont remember the values) . . . .$
	Answered by mathdave last updated on 22/Aug/20

	$m y s o l u t i o n h e r e i t g o e s$ $I = \int_{0}^{\frac{1}{2}} \frac{\ln (1 - x)}{(2 x^{2} - 2 x + 1)} d x = \frac{2}{2} \int_{0}^{\frac{1}{2}} \frac{\ln (1 - x)}{(2 x^{2} - 2 x + 1)} d x = 2 \int_{0}^{\frac{1}{2}} \frac{\ln (1 - x)}{(4 x^{2} - 4 x + 2)} d x$ $I = 2 \int_{0}^{\frac{1}{2}} \frac{\ln (1 - x)}{{(2 x - 1)}^{2} + 1} d x (l e t 2 x - 1 = y a n d d x = \frac{d y}{2})$ $I = \frac{2}{2} \int_{- 1}^{0} \frac{\ln (1 - (\frac{1 + y}{2}))}{y^{2} + 1} d y = \int_{- 1}^{0} \frac{\ln (\frac{1 - y}{2})}{1 + y^{2}} d y$ $l e t y = - t d y = - d t$ $I = \int_{1}^{0} \frac{\ln (\frac{1 + t}{2})}{1 + t^{2}} \times - d t = \int_{0}^{1} \frac{\ln (1 + t)}{1 + t^{2}} d t - \int_{0}^{1} \frac{\ln 2}{1 + t^{2}} d t$ $I = \int_{0}^{1} \frac{\ln (1 + t)}{1 + t^{2}} d t - \ln 2 {[\tan^{- 1} (t)]}_{0}^{1} = \int_{0}^{1} \frac{\ln (1 + t)}{1 + t^{2}} d t - \frac{π}{4} \ln 2$ $l e t \tan x = t a n d d t = se c^{2} x d x a n d$ $1 + \tan^{2} x = \sec^{2} x$ $I = \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \tan x)}{\sec^{2} x} \times \sec^{2} x d x - \frac{π}{4} \ln 2 = \int_{0}^{\frac{π}{4}} \ln (1 + \frac{\sin x}{\cos x}) d x - \frac{π}{4} \ln 2$ $I = \int_{0}^{\frac{π}{4}} \ln (\frac{\sin x + \cos x}{\cos x}) d x - \frac{π}{4} \ln 2 = \int_{0}^{\frac{π}{4}} \ln (\cos x + \sin x) d x - \int_{0}^{\frac{π}{4}} \ln (\cos x) d x - \frac{π}{4} \ln 2$ $s i n c e \int_{0}^{\frac{π}{2}} \ln (\cos x + \sin x) d x = G - \frac{π}{4} \ln 2$ $∵ \int_{0}^{\frac{π}{4}} \ln (\cos x + \sin x) d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\cos x + \sin x) d x = \frac{1}{2} (G - \frac{π}{4} \ln 2)$ $a n d \int_{0}^{\frac{π}{4}} \ln (\cos x) d x = \frac{G}{2} - \frac{π}{4} \ln 2 w h e n u d i n g t h i s$ $i d e n t i t y o f$ $\ln (\cos x) = - \ln 2 + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} \cos (2 n x)}{n} f r o m$ $f o u r i e r s e r i e s$ $∵ I = \frac{G}{2} - \frac{π}{8} \ln 2 - [\frac{G}{2} - \frac{π}{4} \ln 2] - \frac{π}{4} \ln 2$ $I = \frac{G}{2} - \frac{G}{2} - \frac{π}{8} \ln 2 + \frac{π}{4} \ln 2 - \frac{π}{4} \ln 2 = - \frac{π}{8} \ln 2$ $∵ \int_{0}^{\frac{1}{2}} \frac{\ln (1 - x)}{2 x^{2} - 2 x + 1} d x = - \frac{π}{8} \ln 2$ $b y m a t h d a v e (22 / 08 / 2020)$
	Commented by mnjuly1970 last updated on 23/Aug/20

	$s i r . t i k a l l a m o b a r i n a m t i$ $f a r g h s a r m a n b e s h a s h a m . . .$
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