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Question Number 66062 by mathmax by abdo last updated on 08/Aug/19

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

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

2) d e t e r m i n e g (x) = \int_{0}^{1} \frac{d t}{{(c h (t) + x s h (t))}^{2}}

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

Commented by mathmax by abdo last updated on 11/Aug/19

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

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

c h a n g e m e n t e^{t} = u g i v e f (x) = \int_{1}^{e} \frac{2}{(1 + x) u + (1 - x) u^{- 1}} \frac{d u}{u}

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

c a s e 1 \frac{1 - x}{1 + x} > 0 \Rightarrow∣ x ∣< 1 w e d o t h e c h a n g e m e n t u = \sqrt{\frac{1 - x}{1 + x}} z \Rightarrow

f (x) = \frac{2}{1 + x} \int_{\sqrt{\frac{1 + x}{1 - x}}}^{e \sqrt{\frac{1 + x}{1 - x}}} \frac{1 + x}{1 - x} \frac{1}{1 + z^{2}} \sqrt{\frac{1 - x}{1 + x}} d z

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

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

c a s e 2 \frac{1 - x}{1 - x} < 0 \Rightarrow∣ x ∣> 1 \Rightarrow f (x) = \frac{2}{x + 1} \int_{1}^{e} \frac{d u}{u^{2} - {(\sqrt{\frac{x - 1}{x + 1}})}^{2}}

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

= \frac{1}{x - 1} \frac{\sqrt{x - 1}}{\sqrt{x + 1}} \int_{\sqrt{\frac{x + 1}{x - 1}}}^{e \sqrt{\frac{x + 1}{x - 1}}} {\frac{1}{z - 1} - \frac{1}{z + 1}} d z

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

- l n ∣ \frac{\frac{\sqrt{x + 1}}{\sqrt{x - 1}} - 1}{\frac{\sqrt{x + 1}}{\sqrt{x - 1}} + 1} ∣} \Rightarrow

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

Commented by mathmax by abdo last updated on 11/Aug/19

2) s o r r y g (x) = \int_{0}^{1} \frac{s h (t) d t}{{(c h (t) + x s h (t))}^{2}}

w e h a v e f^{^{'}} (x) = - \int_{1}^{1} \frac{s h (t)}{{(c h (t) + x s h (t))}^{2}} = - g (x) \Rightarrow

g (x) = - f^{^{'}} (x) r e s t t o c a l c u l a t e f^{^{'}} (x)

Commented by mathmax by abdo last updated on 11/Aug/19

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