calculate ∫_0 ^(π/4) x artan(2x+1)dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 35685 by prof Abdo imad last updated on 22/May/18

$c a l c u l a t e \int_{0}^{\frac{π}{4}} x a r t a n (2 x + 1) d x$
Commented by prof Abdo imad last updated on 23/May/18
$let put I = ∫_0 ^(π/4) x arctan(2x+1)dx by parts I = [ (x^2 /2)arctan(2x+1)]_0 ^(π/4) − ∫_0 ^(π/4) (x^2 /2) (2/(1 +(2x+1)^2 ))dx =(π^2 /(32))arctan((π/2)+1) −∫_0 ^(π/4) (x^2 /(1+(2x+1)^2 ))dx let calculate J= ∫_0 ^(π/4) (x^2 /(1+(2x+1)^2 ))dx J =_(2x+1=u) ∫_1 ^((π/2)+1) (((((u−1)/2))^2 )/(1+u^2 )) (du/2) = (1/8) ∫_1 ^((π/2)+1) ((u^2 −2u +1)/(1+u^2 ))du 8J = ∫_1 ^((π/2)+1) (1 −((2u)/(1+u^2 )))du = (π/2) −[ ln(1+u^2 )]_1 ^((π/2)+1) =(π/2) −{ ln(1+((π/2)+1)^2 ) −ln(2)} ⇒ J = (π/(16)) −(1/8){ln(1+((π/2)+1)^2 ) −ln(2)} so I = (π^2 /(32)) arctan((π/2)+1) −(π/(16)) +(1/8){ln(1+((π/2)+1)^2 )−ln(2)}$
$l e t p u t I = \int_{0}^{\frac{π}{4}} x a r c t a n (2 x + 1) d x b y p a r t s$ $I = {[\frac{x^{2}}{2} a r c t a n (2 x + 1)]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} \frac{x^{2}}{2} \frac{2}{1 + {(2 x + 1)}^{2}} d x$ $= \frac{π^{2}}{32} a r c t a n (\frac{π}{2} + 1) - \int_{0}^{\frac{π}{4}} \frac{x^{2}}{1 + {(2 x + 1)}^{2}} d x l e t$ $c a l c u l a t e J = \int_{0}^{\frac{π}{4}} \frac{x^{2}}{1 + {(2 x + 1)}^{2}} d x$ $J =_{2 x + 1 = u} \int_{1}^{\frac{π}{2} + 1} \frac{{(\frac{u - 1}{2})}^{2}}{1 + u^{2}} \frac{d u}{2}$ $= \frac{1}{8} \int_{1}^{\frac{π}{2} + 1} \frac{u^{2} - 2 u + 1}{1 + u^{2}} d u$ $8 J = \int_{1}^{\frac{π}{2} + 1} (1 - \frac{2 u}{1 + u^{2}}) d u$ $= \frac{π}{2} - {[l n (1 + u^{2})]}_{1}^{\frac{π}{2} + 1}$ $= \frac{π}{2} - {l n (1 + {(\frac{π}{2} + 1)}^{2}) - l n (2)} \Rightarrow$ $J = \frac{π}{16} - \frac{1}{8} {l n (1 + {(\frac{π}{2} + 1)}^{2}) - l n (2)} s o$ $I = \frac{π^{2}}{32} a r c t a n (\frac{π}{2} + 1) - \frac{π}{16} + \frac{1}{8} {l n (1 + {(\frac{π}{2} + 1)}^{2}) - l n (2)}$
	Answered by ajfour last updated on 23/May/18

	$I = \frac{x^{2}}{2} \tan^{- 1} (2 x + 1) ∣_{0}^{π / 4} - \int_{0}^{π / 4} (\frac{x^{2}}{2}) [\frac{2}{1 + {(2 x + 1)}^{2}}] d x$ $= \frac{π^{2}}{32} \tan^{- 1} (1 + \frac{π}{2}) - I_{1}$ $I_{1} = \frac{1}{4} \int_{0}^{π / 4} \frac{1 + {(2 x + 1)}^{2} - 2 (2 x + 1)}{1 + {(2 x + 1)}^{2}} d x$ $= \frac{x}{4} ∣_{0}^{π / 4} - \frac{1}{4} \int_{2}^{1 + {(1 + \frac{π}{2})}^{2}} \frac{d t}{t}$ $= \frac{π}{16} - \frac{1}{4} \ln [\frac{1}{2} + \frac{{(π + 2)}^{2}}{8}]$ $I = \frac{π^{2}}{32} \tan^{- 1} (1 + \frac{π}{2}) - \frac{π}{16}$ $+ \frac{1}{4} \ln [\frac{1}{2} + \frac{{(π + 2)}^{2}}{8}] .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum