calculate ∫_0 ^1 (dx/(2sh(x)+3ch(x)))


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Question Number 59526 by Mr X pcx last updated on 11/May/19

$c a l c u l a t e \int_{0}^{1} \frac{d x}{2 s h (x) + 3 c h (x)}$
Commented by maxmathsup by imad last updated on 12/May/19
$let I =∫_0 ^1 (dx/(2sh(x)+3ch(x))) ⇒I =∫_0 ^1 (dx/(2((e^x −e^(−x) )/2) +3((e^x +e^(−x) )/2))) =∫_0 ^1 ((2dx)/(2e^x −2e^(−x) +3e^x +3e^(−x) )) =2∫_0 ^1 (dx/(5e^x +e^(−x) )) =_(e^x =t) 2 ∫_1 ^e (dt/(t(5t +(1/t)))) =2 ∫_1 ^e (dt/(5t^2 +1)) =_((√5)t =u) 2 ∫_(√5) ^(e(√5)) (du/((√5)(u^2 +1))) =(2/(√5)) [arctan(u)]_(√5) ^(e(√5)) =(2/(√5)){arctan(e(√5))−arctan((√5))}$
$l e t I = \int_{0}^{1} \frac{d x}{2 s h (x) + 3 c h (x)} \Rightarrow I = \int_{0}^{1} \frac{d x}{2 \frac{e^{x} - e^{- x}}{2} + 3 \frac{e^{x} + e^{- x}}{2}}$ $= \int_{0}^{1} \frac{2 d x}{2 e^{x} - 2 e^{- x} + 3 e^{x} + 3 e^{- x}} = 2 \int_{0}^{1} \frac{d x}{5 e^{x} + e^{- x}} =_{e^{x} = t} 2 \int_{1}^{e} \frac{d t}{t (5 t + \frac{1}{t})}$ $= 2 \int_{1}^{e} \frac{d t}{5 t^{2} + 1} =_{\sqrt{5} t = u} 2 \int_{\sqrt{5}}^{e \sqrt{5}} \frac{d u}{\sqrt{5} (u^{2} + 1)} = \frac{2}{\sqrt{5}} {[a r c t a n (u)]}_{\sqrt{5}}^{e \sqrt{5}}$ $= \frac{2}{\sqrt{5}} {a r c t a n (e \sqrt{5}) - a r c t a n (\sqrt{5})}$
	Answered by MJS last updated on 11/May/19

	$\int \frac{d x}{2 \sinh x + 3 \cosh x} = 2 \int \frac{e^{x}}{{5 e}^{2 x} + 1} d x =$ $[t = \sqrt{5} e^{x} \to d x = \frac{d t}{t}]$ $= \frac{2 \sqrt{5}}{5} \int \frac{d t}{t^{2} + 1} = \frac{2 \sqrt{5}}{5} \arctan t = \frac{2 \sqrt{5}}{5} \arctan \sqrt{5} e^{x} + C$ $\underset{0}{\int^{1}} \frac{d x}{2 \sinh x + 3 \cosh x} = \frac{2 \sqrt{5}}{5} (\arctan \sqrt{5} e - \arctan \sqrt{5}) \approx . 230292$
	Commented by maxmathsup by imad last updated on 13/May/19

	$t h a n k s s i r m j s .$
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