The-integral-0-1-2-ln-1-2x-1-4x-2-dx-a-pi-4-ln2-b-pi-8-ln2-c-pi-16-ln2-d-pi-32-ln2-

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 \dots

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 = 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