let-f-x-0-1-dt-1-xch-t-with-x-real-1-determine-a-explicit-form-of-f-x-2-find-also-g-x-0-1-dt-1-xch-t-2-3-calculate-0-1-dt-1-3ch-t-and-0-1-dt-1-3ch-t-

Question Number 59528 by Mr X pcx last updated on 11/May/19

l e t f (x) = \int_{0}^{1} \frac{d t}{1 + x c h (t)} w i t h x r e a l

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

2) f i n d a l s o g (x) = \int_{0}^{1} \frac{d t}{{(1 + x c h (t))}^{2}}

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

\int_{0}^{1} \frac{d t}{{(1 + 3 c h (t))}^{2}}

Commented by MJS last updated on 11/May/19

same path as before

\int \frac{d t}{1 + x \cosh t} = 2 \int \frac{e^{t}}{x e^{2 t} + {2 e}^{t} + x} d t = \frac{2}{\sqrt{x^{2} - 1}} \arctan \frac{x e^{t} + 1}{\sqrt{x^{2} - 1}} + C

\int \frac{d t}{{(1 + x \cosh t)}^{2}} = 4 \int \frac{e^{2 t}}{{(x e^{2 t} + {2 e}^{t} + x)}^{2}} d t = - \frac{2}{{(x^{2} - 1)}^{3 / 2}} \arctan \frac{x e^{t} + 1}{\sqrt{x^{2} - 1}} - \frac{2 (e^{t} + x)}{(x e^{2 t} + {2 e}^{t} + x) (x^{2} - 1)}

Commented by maxmathsup by imad last updated on 11/May/19

t h a n k s s i r

Commented by maxmathsup by imad last updated on 12/May/19

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

=_{e^{t} = u} \int_{1}^{e} \frac{2}{2 + x u + {x u}^{- 1}} \frac{d u}{u} = \int_{1}^{e} \frac{2 d u}{2 u + {x u}^{2} + x}

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

{x u}^{2} + 2 u + x = 0 \to Δ^{^{'}} = 1 - x^{2} c a s e 1 1 - x^{2} > 0 \Rightarrow∣ x ∣< 1 \Rightarrow u_{1} = \frac{- 1 + \sqrt{1 - x^{2}}}{x}

u_{2} = \frac{- 1 - \sqrt{1 - x^{2}}}{x} (w e s u p p o s e x \neq 0) \Rightarrow F (u) = \frac{2}{x (u - u_{1}) (u - u_{2})}

= \frac{2}{x (u_{1} - u_{2})} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) = \frac{2}{x \frac{2 \sqrt{1 - x^{2}}}{x}} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) = \frac{1}{\sqrt{1 - x^{2}}} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}})

f (x) = \int_{1}^{e} F (u) d u = \frac{1}{\sqrt{1 - x^{2}}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{1}^{e} = \frac{1}{\sqrt{1 - x^{2}}} {l n ∣ \frac{e - u_{1}}{e - u_{2}} ∣ - l n ∣ \frac{1 - u_{1}}{1 - u_{2}} ∣}

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

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

Commented by maxmathsup by imad last updated on 12/May/19

c a s e 2 1 - x^{2} < 0 \Rightarrow x^{2} - 1 > 0 \Rightarrow F (u) = \frac{2}{{x u}^{2} + 2 u + x} = \frac{2}{x (u^{2} + 2 \frac{u}{x} + 1)}

= \frac{2}{x {u^{2} + 2 \frac{u}{x} + \frac{1}{x^{2}} + 1 - \frac{1}{x^{2}})} = \frac{2}{x {{(u + \frac{1}{x})}^{2} + \frac{x^{2} - 1}{x^{2}}}}

= \frac{2}{x {{(u + \frac{1}{x})}^{2} + {(\frac{\sqrt{x^{2} - 1}}{∣ x ∣})}^{2}} \Rightarrow c h a n g . u + \frac{1}{x} = \frac{\sqrt{x^{2} - 1}}{∣ x ∣} α g i v e α = \frac{x u + 1}{x} \frac{∣ x ∣}{\sqrt{x^{2} - 1}}

f (x) = \int_{1}^{e} F (u) d u = \frac{2}{x} \int_{ξ (x) \frac{x + 1}{\sqrt{x^{2} - 1}}}^{ξ (x) \frac{e x + 1}{\sqrt{x^{2} - 1}}} \frac{1}{\frac{x^{2} - 1}{∣ x ∣} (1 + α^{2})} \frac{\sqrt{x^{2} - 1}}{∣ x ∣} d α

= \frac{2}{x \sqrt{x^{2} - 1}} {[a r c t a n (α)]}_{\frac{x + 1}{\sqrt{x^{2} - 1}} ξ (x)}^{\frac{e x + 1}{\sqrt{x^{2} - 1}} ξ (x)}

= \frac{2}{x \sqrt{x^{2} - 1}} {a r c t a n (\frac{e x + 1}{\sqrt{x^{2} - 1}} ξ (x)) - a r c t a n (\frac{x + 1}{\sqrt{x^{2} - 1}} ξ (x))} w i t h

ξ (x) = 1 i f x > 0 a n d ξ (x) = - 1 i f x < 0 .

Commented by maxmathsup by imad last updated on 12/May/19

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

= \frac{2}{3 \sqrt{8}} {a r c t a n (\frac{3 e + 1}{\sqrt{8}}) - a r c t a n (\frac{4}{\sqrt{8}})}

= \frac{1}{3 \sqrt{2}} {a r c t a n (\frac{3 e + 1}{2 \sqrt{2}}) - a r c t a n (\sqrt{2})} .