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

Question Number 209217 by mnjuly1970 last updated on 04/Jul/24

c a l c u l a t e :

I = \int_{0}^{\infty} \frac{{t a n}^{- 1} (x)}{{(1 + x^{2})}^{2}} d x = ?

Answered by Berbere last updated on 04/Jul/24

\tan^{- 1} (x) = u

\int_{0}^{\frac{π}{2}} {u c o s}^{2} (u) d u = \int_{0}^{\frac{π}{2}} u . (\frac{1 + c o s (2 u)}{2})

= \frac{1}{4} . \frac{π^{2}}{4} + \frac{1}{2} {[\frac{s i n (2 u)}{2} u]}_{0}^{\frac{π}{2}} - \frac{1}{4} \int s i n (2 u) d u

= \frac{π^{2}}{16} + \frac{1}{8} {[c o s (2 u)]}_{0}^{\frac{π}{2}} = \frac{π^{2}}{16} - \frac{1}{4}

Answered by lepuissantcedricjunior last updated on 05/Jul/24

i = \int_{0}^{\infty} \frac{a r c t a n (x)}{{(1 + x^{2})}^{2}} d x

p o s o n s x = t a n t => d x = (1 + {t a n}^{2} t) d t

i = \int_{0}^{π / 2} \frac{t}{1 + {t a n}^{2} t} d t = \int_{0}^{\frac{π}{2}} {t c o s}^{2} (t) d t

{\begin{cases} u = t \\ v^{'} = {c o s}^{2} (t) \end{cases} => {\begin{cases} u^{'} = 1 \\ v = \frac{t}{2} + \frac{s i n 2 t}{4} \end{cases}

i = {[\frac{t^{2}}{2} + \frac{t s i n 2 t}{4}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} (\frac{t}{2} + \frac{s i n 2 t}{4}) d t

= \frac{π^{2}}{8} - {[\frac{t^{2}}{4} - \frac{c o s 2 t}{8}]}_{0}^{\frac{π}{2}}

= \frac{π^{2}}{8} - \frac{π^{2}}{16} - \frac{1}{8} - \frac{1}{8} = \frac{π^{2}}{16} - \frac{1}{4}

\Rightarrow i = \int_{0}^{\infty} \frac{a r c t a n (x)}{{(1 + x^{2})}^{2}} d x = \frac{π^{2}}{16} - \frac{1}{4}