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Question Number 72111 by A8;15: last updated on 24/Oct/19

Commented by mathmax by abdo last updated on 24/Oct/19
$let I=∫_0 ^(e/π) ((arctan(((πx)/e)))/(πx +e))dx changement ((πx)/e) =t give I =∫_0 ^1 ((arctant)/(π(((et)/π))+e))(e/π)dt =(e/π)∫_0 ^1 ((arctan(t))/(e(t+1)))dt =(1/π) ∫_0 ^1 ((arctan(t))/(t+1))dt by parts u^′ =(1/(t+1)) and v=arctan(t) ⇒ ∫_0 ^1 ((arctan(t))/(t+1))dt =[ln(t+1)arctant]_0 ^1 −∫_0 ^1 ((ln(t+1))/(1+t^2 ))dt =(π/4)ln(2)−∫_0 ^1 ((ln(t+1))/(1+t^2 ))dt but ∫_0 ^1 ((ln(t+1))/(1+t^2 ))dt =_(t=tanθ) ∫_0 ^(π/4) ((ln(tanθ +1))/(1+tan^2 θ))(1+tan^2 θ)dθ =∫_0 ^(π/4) ln(((sinθ)/(cosθ))+1)dθ =∫_0 ^(π/4) ln(cosθ +sinθ)−ln(cosθ)}dθ =∫_0 ^(π/4) ln((√2)cos((π/4)−θ))−∫_0 ^(π/4) ln(cosθ)dθ =_((π/4)−θ=u) −∫_0 ^(π/4) { ((1/2)(π/4))ln(2)+ln(cosu)}(−du)−∫_0 ^(π/4) ln(cosu)du =−(π/8)ln(2) ⇒∫_0 ^1 ((arctan(t))/(1+t^2 ))dt =(π/4)ln2−(π/8)ln(2) ⇒I=((ln(2))/8)$
$l e t I = \int_{0}^{\frac{e}{π}} \frac{a r c t a n (\frac{π x}{e})}{π x + e} d x c h a n g e m e n t \frac{π x}{e} = t g i v e$ $I = \int_{0}^{1} \frac{a r c t a n t}{π (\frac{e t}{π}) + e} \frac{e}{π} d t = \frac{e}{π} \int_{0}^{1} \frac{a r c t a n (t)}{e (t + 1)} d t = \frac{1}{π} \int_{0}^{1} \frac{a r c t a n (t)}{t + 1} d t$ $b y p a r t s u^{^{'}} = \frac{1}{t + 1} a n d v = a r c t a n (t) \Rightarrow$ $\int_{0}^{1} \frac{a r c t a n (t)}{t + 1} d t = {[l n (t + 1) a r c t a n t]}_{0}^{1} - \int_{0}^{1} \frac{l n (t + 1)}{1 + t^{2}} d t$ $= \frac{π}{4} l n (2) - \int_{0}^{1} \frac{l n (t + 1)}{1 + t^{2}} d t b u t$ $\int_{0}^{1} \frac{l n (t + 1)}{1 + t^{2}} d t =_{t = t a n θ} \int_{0}^{\frac{π}{4}} \frac{l n (t a n θ + 1)}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) d θ$ $= \int_{0}^{\frac{π}{4}} l n (\frac{s i n θ}{c o s θ} + 1) d θ = \int_{0}^{\frac{π}{4}} l n (c o s θ + s i n θ) - l n (c o s θ)} d θ$ $= \int_{0}^{\frac{π}{4}} l n (\sqrt{2} c o s (\frac{π}{4} - θ)) - \int_{0}^{\frac{π}{4}} l n (c o s θ) d θ$ $=_{\frac{π}{4} - θ = u} - \int_{0}^{\frac{π}{4}} {(\frac{1}{2} \frac{π}{4}) l n (2) + l n (c o s u)} (- d u) - \int_{0}^{\frac{π}{4}} l n (c o s u) d u$ $= - \frac{π}{8} l n (2) \Rightarrow \int_{0}^{1} \frac{a r c t a n (t)}{1 + t^{2}} d t = \frac{π}{4} l n 2 - \frac{π}{8} l n (2) \Rightarrow I = \frac{l n (2)}{8}$
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