The integral ∫_0 ^(1/2) ((ln (1+2x))/(1+4x^2 ))dx = ? a) (π/4)ln2 b)(π/8)ln2 c)(π/(16))ln2 d)(π/(32))ln2


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Question Number 49827 by rahul 19 last updated on 11/Dec/18

$T h e i n t e g r a l \int_{0}^{\frac{1}{2}} \frac{\ln (1 + 2 x)}{1 + 4 x^{2}} d x = ?$ $a) \frac{π}{4} l n 2 b) \frac{π}{8} l n 2 c) \frac{π}{16} l n 2 d) \frac{π}{32} l n 2$
Commented by rahul 19 last updated on 11/Dec/18

$t h a n k y o u s i r!$
Commented by rahul 19 last updated on 11/Dec/18

$A n y s h o r t / t r i c k y m e t h o d o t h e r$ $t h a n u s u a l s u b s t i t u t i o n : 2 x = \tan θ ? ? ?$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

$r e m e m b e r t h e a n s w e r . . .$
Commented by Abdo msup. last updated on 11/Dec/18

$l e t I = \int_{0}^{\frac{1}{2}} \frac{l n (1 + 2 x)}{1 + 4 x^{2}} d x c h a n g e m e n t 2 x = t a n t g i v e$ $I = \frac{1}{2} \int_{0}^{\frac{π}{4}} \frac{l n (1 + t a n t)}{1 + {t a n}^{2} t} (1 + {t a n}^{2} t) d t \Rightarrow$ $2 I = \int_{0}^{\frac{π}{4}} l n (1 + t a n t) d t = \int_{0}^{\frac{π}{4}} l n (\frac{c o s t + s i n t}{c o s t}) d t$ $= \int_{0}^{\frac{π}{4}} l n (c o s t + s i n t) d t - \int_{0}^{\frac{π}{4}} l n (c o s t) d t$ $= \int_{0}^{\frac{π}{4}} l n (\sqrt{2} s i n (t + \frac{π}{4}) d t - \int_{0}^{\frac{π}{4}} l n (c o s t) d t$ $= \frac{π}{8} l n (2) + \int_{0}^{\frac{π}{4}} l n (s i n (t + \frac{π}{4})) d t - \int_{0}^{\frac{π}{4}} l n (c o s t) d t$ $\int_{0}^{\frac{π}{4}} l n (s i n (t + \frac{π}{4})) d t =_{t + \frac{π}{4} = u} \int_{\frac{π}{4}}^{\frac{π}{2}} l n (s i n u) d u$ $=_{u = \frac{π}{2} - α} \int_{\frac{π}{4}}^{0} l n (c o s α) (- d α) = \int_{0}^{\frac{π}{4}} l n (c o s α) d α$ $2 I = \frac{π}{8} l n (2) \Rightarrow I = \frac{π}{16} l n (2) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18
	$t=2x dt=2dx ∫_0 ^1 ((ln(1+t))/(1+t^2 ))dt I=∫_0 ^1 ((ln(1+t))/(1+t^2 ))dt t=tana dt=sec^2 ada I=∫_0 ^(π/4) ((ln(1+tana))/(sec^2 a))×sec^2 ada =∫_0 ^(π/4) ln{1+tan((π/4)−a)} =∫_0 ^(π/4) ln{1+((1−tana)/(1+tana))}da =∫_0 ^(π/4) ln((2/(1+tana)))da =∫_0 ^(π/4) ln2 da−∫_0 ^(π/4) ln(1+tana)da 2I=ln2∫_0 ^(π/4) da I=(1/2)×(ln2)×(π/4)=(π/8)ln2$
	$t = 2 x d t = 2 d x$ $\int_{0}^{1} \frac{l n (1 + t)}{1 + t^{2}} d t$ $I = \int_{0}^{1} \frac{l n (1 + t)}{1 + t^{2}} d t$ $t = t a n a d t = {s e c}^{2} a d a$ $I = \int_{0}^{\frac{π}{4}} \frac{l n (1 + t a n a)}{{s e c}^{2} a} \times {s e c}^{2} a d a$ $= \int_{0}^{\frac{π}{4}} l n {1 + t a n (\frac{π}{4} - a)}$ $= \int_{0}^{\frac{π}{4}} l n {1 + \frac{1 - t a n a}{1 + t a n a}} d a$ $= \int_{0}^{\frac{π}{4}} l n (\frac{2}{1 + t a n a}) d a$ $= \int_{0}^{\frac{π}{4}} l n 2 d a - \int_{0}^{\frac{π}{4}} l n (1 + t a n a) d a$ $2 I = l n 2 \int_{0}^{\frac{π}{4}} d a$ $I = \frac{1}{2} \times (l n 2) \times \frac{π}{4} = \frac{π}{8} l n 2$
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