find-the-value-of-I-0-1-arctan-2x-1-4x-2-dx-

Question Number 39024 by maxmathsup by imad last updated on 01/Jul/18

f i n d t h e v a l u e o f I = \int_{0}^{1} \frac{a r c t a n (2 x)}{\sqrt{1 + 4 x^{2}}} d x

Commented by math khazana by abdo last updated on 04/Jul/18

t h e Q i s f i n d \int_{0}^{1} \frac{a r c t a n (2 x)}{1 + 4 x^{2}} d x .

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

l e t I = \int_{0}^{1} \frac{a r c t a n (2 x)}{1 + 4 x^{2}} l e t i n t e g r a t e b y p a r t s

u^{^{'}} = \frac{1}{1 + 4 x^{2}} a n d v = a r c t a n (2 x) \Rightarrow

I = {[\frac{1}{2} {(a r c t a n (2 x))}^{2}]}_{0}^{1} - \int_{0}^{1} \frac{1}{2} a r c t a n (2 x) \frac{2}{1 + 4 x^{2}} d x

= \frac{1}{2} {(a r c t a n (2))}^{2} - I \Rightarrow

2 I = \frac{1}{2} {a r c t a n (2)}^{2} \Rightarrow I = \frac{1}{4} {a r c t a n (2)}^{2}