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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

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}] .

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