find-0-1-arctan-x-1-x-2-dx-

Question Number 48667 by maxmathsup by imad last updated on 26/Nov/18

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

Commented by Abdo msup. last updated on 02/Dec/18

c h a n g e m e n t x = t a n t g i v e

I = \int_{0}^{\frac{π}{4}} \frac{t}{\sqrt{1 + {t a n}^{2} t}} (1 + {t a n}^{2} t) d t = \int_{0}^{\frac{π}{4}} t \frac{d t}{c o s t}

= \int_{0}^{\frac{π}{4}} \frac{t}{c o s t} d t =_{t a n (\frac{t}{2}) = u} \int_{0}^{\sqrt{2} - 1} \frac{2 a r c t a n u}{\frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= 4 \int_{0}^{\sqrt{2} - 1} \frac{a r c t a n (u)}{1 - u^{2}} d u = 4 \int_{0}^{\sqrt{2} - 1} a r c t a n (u) (\sum_{n = 0}^{\infty} u^{2 n}) d u

= 4 \sum_{n = 0}^{\infty} \int_{0}^{\sqrt{2} - 1} u^{2 n} a r c t a n u d u = 4 \sum_{n = 0}^{\infty} A_{n} b y p a r t s

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

+ \int_{0}^{\sqrt{2} - 1} \frac{1}{2 n + 1} u^{2 n + 1} \frac{d u}{1 + u^{2}}

= \frac{1}{2 n + 1} {(\sqrt{2} - 1)}^{2 n + 1} a r c t a n (\sqrt{2} - 1) + \frac{1}{2 n + 1} \int_{0}^{\sqrt{2} - 1} \frac{u^{2 n + 1}}{1 + u^{2}} d u

= \frac{π {(\sqrt{2} - 1)}^{2 n + 1}}{8 (2 n + 1)} + \frac{1}{2 n + 1} \int_{0}^{\sqrt{2} - 1} \frac{u^{2 n + 1}}{1 + u^{2}} d u b u t

\int_{0}^{\sqrt{2} - 1} \frac{u^{2 n + 1}}{1 + u^{2}} d u =_{t a n θ = u} \int_{0}^{\frac{π}{8}} \frac{{t a n}^{2 n + 1} θ}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) d θ

= \int_{0}^{\frac{π}{8}} {t a n}^{2 n + 1} θ d θ \dots . b e c o n t i n u e d \dots .

Answered by Abdulhafeez Abu qatada last updated on 26/Nov/18