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Question Number 42086 by maxmathsup by imad last updated on 17/Aug/18

l e t f (x) = \int_{0}^{2} \frac{c h (t)}{2 x s h (t) + 1} d t

1) f i n d a s i m p l e f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{2} \frac{c h (t)}{1 + s h (t)} d t

3) c a l c u l a t e \int_{0}^{2} \frac{c h (t)}{3 s h (t) + 1} d t .

Answered by maxmathsup by imad last updated on 18/Aug/18

1) w e h a v e f (x) = \int_{0}^{2} \frac{\frac{e^{t} + e^{- t}}{2}}{2 x \frac{e^{t} - e^{- t}}{2} + 1} d t = \frac{1}{2} \int_{0}^{2} \frac{e^{t} + e^{- t}}{{x e}^{t} - x e^{- t} + 1} d t

= \frac{1}{2} \int_{0}^{2} \frac{e^{2 t} + 1}{x e^{z t} - x + e t} d t \Rightarrow 2 f (x) =_{e^{t} = u} \int_{1}^{e^{2}} \frac{u^{2} + 1}{{x u}^{2} - x + u} \frac{d u}{u}

= \int_{1}^{e^{2}} \frac{u^{2} + 1}{u ({x u}^{2} + u - x)} d u l e t d e c o m p o s e F (u) = \frac{u^{2} + 1}{u ({x u}^{2} + u - x)}

p o l e s o f F

Δ = 1 - 4 (- x^{2}) = 1 + 4 x^{2} \Rightarrow u_{1} = \frac{- 1 + \sqrt{4 x^{2} + 1}}{2 x}

u_{2} = \frac{- 1 - \sqrt{4 x^{2} + 1}}{2 x} (x \neq 0) \Rightarrow

F (u) = \frac{u^{2} + 1}{x u (u - u_{1}) (u - u_{2})} = \frac{a}{u} + \frac{b}{u - u_{1}} + \frac{c}{u - u_{2}}

a = {l i m}_{u \to 0} u F (u) = \frac{1}{x u_{1} u_{2}} = - \frac{1}{x}

b = {l i m}_{u \to u_{1}} (u - u_{1}) F (u) = \frac{u_{1}^{2} + 1}{{x u}_{1} (u_{1} - u_{2})}

c = {l i m}_{u \to u_{2}} (u - u_{2}) F (u) = \frac{u_{2}^{2} + 1}{{x u}_{2} (u_{2} - u_{1})} \Rightarrow

f (x) = \int_{1}^{e^{2}} F (u) d u = a \int_{1}^{e^{2}} \frac{d u}{u} + b \int_{1}^{e^{2}} \frac{d u}{u - u_{1}} + c \int_{1}^{e^{2}} \frac{d u}{u - u_{2}}

= a {[l n ∣ u ∣]}_{1}^{e^{2}} + b {[l n ∣ u - u_{1} ∣]}_{1}^{e^{2}} + c {[l n ∣ u - u_{2} ∣]}_{1}^{e^{2}}

= 2 a + b (l n ∣ \frac{e^{2} - u_{1}}{1 - u_{1}} ∣) + c l n ∣ \frac{e^{2} - u_{2}}{1 - u_{2}} ∣

f (x) = \frac{- 2}{x} + \frac{{(\frac{- 1 + \sqrt{4 x^{2} + 1}}{2 x})}^{2} + 1}{x (\frac{- 1 + \sqrt{4 x^{2} + 1}}{2 x}) (\frac{\sqrt{4 x^{2} + 1}}{x})} l n ∣ \frac{e^{2} - \frac{- 1 + \sqrt{4 x^{2} + 1}}{2 x}}{1 - \frac{- 1 + \sqrt{4 x^{2} + 1}}{2 x}} ∣

+ \frac{{(\frac{1 + \sqrt{4 x^{2} + 1}}{2 x})}^{2} + 1}{x (\frac{- 1 - \sqrt{4 x^{2} + 1}}{2 x}) (- \frac{\sqrt{4 x^{2} + 1}}{x})} l n ∣ \frac{e^{2} + \frac{1 + \sqrt{4 x^{2} + 1}}{2 x}}{1 + \frac{1 + \sqrt{4 x^{2} + 1}}{2 x}} ∣

f (x) = \frac{- 2}{x} + \frac{{(- 1 + \sqrt{4 x^{2} + 1})}^{2} + 4 x^{2}}{2 x} l n ∣ \frac{2 e^{2} x - (- 1 + \sqrt{4 x^{2} + 1})}{2 x - (- 1 + \sqrt{4 x^{2} + 1}} ∣

- \frac{{(1 + \sqrt{4 x^{2} + 1})}^{2} + 4 x^{2}}{2 x} l n ∣ \frac{2 e^{2} x + 1 + \sqrt{4 x^{2} + 1}}{2 x + 1 + \sqrt{4 x^{2} + 1}} ∣

Commented by maxmathsup by imad last updated on 18/Aug/18

2 f (x) = - \frac{2}{x} + \dots . \Rightarrow

f (x) = - \frac{1}{x} + \frac{{(- 1 + \sqrt{4 x^{2} + 1})}^{2} + 4 x^{2}}{4 x} l n ∣ \frac{2 e^{2} x - (- 1 + \sqrt{4 x^{2} + 1}}{2 x - (- 1 + \sqrt{4 x^{2} + 1}} ∣

- \frac{{(1 + \sqrt{4 x^{2} + 1)})}^{2} + 4 x^{2}}{4 x} l n ∣ \frac{2 e^{2} x + 1 + \sqrt{4 x^{2} + 1}}{2 x + 1 + \sqrt{4 x^{2} + 1}} ∣ .

Commented by math khazana by abdo last updated on 18/Aug/18

\int_{0}^{2} \frac{c h (t)}{1 + s h (t)} d t = f (\frac{1}{2})

= - 2 + \frac{{(- 1 + \sqrt{2})}^{2} + 1}{2} l n ∣ \frac{e^{2} - (- 1 + \sqrt{2})}{1 - (- 1 + \sqrt{2})} ∣

- \frac{{(1 + \sqrt{2})}^{2} + 1}{2} l n ∣ \frac{e^{2} + 1 + \sqrt{2}}{2 + \sqrt{2}} ∣

= - 2 + (2 - \sqrt{2}) l n ∣ \frac{e^{2} + 1 - \sqrt{2}}{2 - \sqrt{2}} ∣ - (2 + \sqrt{2}) l n ∣ \frac{e^{2} + 1 + \sqrt{2}}{2 + \sqrt{2}} ∣ .