calculate-I-0-1-x-2-1-x-2-arctan-x-dx-

Question Number 42796 by maxmathsup by imad last updated on 02/Sep/18

c a l c u l a t e I = \int_{0}^{1} \frac{x^{2}}{1 + x^{2}} a r c t a n (x) d x

Commented by maxmathsup by imad last updated on 04/Sep/18

w e h a v e I = \int_{0}^{1} \frac{1 + x^{2} - 1}{1 + x^{2}} a r c t a n (x) d x = \int_{0}^{1} a r c t a n x - \int_{0}^{1} \frac{a r c t a n x}{1 + x^{2}} d x b u t

b y p a r t s \int_{0}^{1} a r c t a n x d x = {[x a r c t a n x]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^{2}} d x

= \frac{π}{4} - {[\frac{1}{2} l n (1 + x^{2})]}_{0}^{1} = \frac{π}{4} - \frac{l n (2)}{2} a l s o b y p a r t s u^{^{'}} = \frac{1}{1 + x^{2}} a n d v = a r c t a n (x)

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

2 \int_{0}^{1} \frac{a r c t a n (x)}{1 + x^{2}} d x = \frac{π^{2}}{16} \Rightarrow \int_{0}^{1} \frac{a r c t a n (x)}{1 + x^{2}} d x = \frac{π^{2}}{32} \Rightarrow

I = \frac{π}{4} - \frac{l n (2)}{2} - \frac{π^{2}}{32} .

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18

t = {t a n}^{- 1} x d t = \frac{d x}{1 + x^{2}}

\int_{0}^{\frac{Π}{4}} \frac{{t a n}^{2} t \times t}{} d t

\int_{0}^{\frac{Π}{4}} ({s e c}^{2} t - 1) t d t

\int_{0}^{\frac{Π}{4}} {t s e c}^{2} t d t - \int_{0}^{\frac{Π}{4}} t d t

∣ t t a n t - l n s e c t - \frac{t^{2}}{2} ∣_{0}^{\frac{Π}{4}}

\frac{Π}{4} - \frac{1}{2} l n 2 - \frac{\prod^{2}}{32}