let-x-gt-0-and-f-x-1-2-t-1-t-2-2xt-1-dt-1-find-a-explicit-form-of-f-x-2-determine-also-g-x-1-2-t-2-t-t-2-2xt-1-dt-3-find-the-value-of-integrals-1-2-t-1-t-2-t-1-dt

Question Number 66816 by mathmax by abdo last updated on 20/Aug/19

l e t x > 0 a n d f (x) = \int_{1}^{2} (t + 1) \sqrt{t^{2} - 2 x t - 1} d 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 a l s o g (x) = \int_{1}^{2} \frac{t^{2} + t}{\sqrt{t^{2} - 2 x t - 1}} d t

3) f i n d t h e v a l u e o f i n t e g r a l s \int_{1}^{2} (t + 1) \sqrt{t^{2} - t - 1} d t

a n d \int_{1}^{2} \frac{t^{2} + t}{\sqrt{t^{2} - t - 1}} d t .

Commented by mathmax by abdo last updated on 23/Aug/19

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

t^{2} - 2 x t - 1 = t^{2} - 2 x t + x^{2} - x^{2} - 1 = {(t - x)}^{2} - {(\sqrt{1 + x^{2}})}^{2}

w e d o t h e c h a n g e m e n t t - x = \sqrt{1 + x^{2}} c h (u) \Rightarrow

f (x) = \int_{a r g c h (\frac{1 - x}{\sqrt{1 + x^{2}}})}^{a r c h (\frac{2 - x}{\sqrt{1 + x^{2}}})} (x + \sqrt{1 + x^{2}} c h (u)) \sqrt{1 + x^{2}} s h (u) d u

= x \sqrt{1 + x^{2}} \int_{a r g c h (\frac{1 - x}{\sqrt{1 + x^{2}}})}^{a r g c h (\frac{2 - x}{\sqrt{1 + x^{2}}})} s h (u) d u + (1 + x^{2}) \int_{a r g c h (\frac{1 - x}{\sqrt{1 + x^{2}}})}^{a r g c h (\frac{2 - x}{\sqrt{1 + x^{2}}})} c h (u) s h (u) d u

w e h a v e \int_{a r g c h (\frac{1 - x}{\sqrt{1 + x^{2}}})}^{a r g c h (\frac{2 - x}{\sqrt{1 + x^{2}}})} s h (u) d u = {[c h (u)]}_{a r g (\dots)}^{a r g (\dots)}

a r g c h (\frac{2 - x}{\sqrt{1 + x^{2}}}) = l n (\frac{2 - x}{\sqrt{1 + x^{2}}} + \sqrt{1 + \frac{{(2 - x)}^{2}}{1 + x^{2}}})

a r g c h (\frac{1 - x}{\sqrt{1 + x^{2}}}) = l n (\frac{1 - x}{\sqrt{1 + x^{2}}} + \sqrt{1 + \frac{{(1 - x)}^{2}}{1 + x^{2}}}) \Rightarrow

\int_{a r g c h (\frac{1 - x}{\sqrt{1 + x^{2}}})}^{a r g c h (\frac{2 - x}{\sqrt{1 + x^{2}}})} s h (u) d u = \frac{1}{2} [e^{u} - e^{- u}] . . = \frac{1}{2} {\frac{2 - x}{\sqrt{1 + x^{2}}} + \sqrt{1 + \frac{{(2 - x)}^{2}}{1 + x^{2}}}

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

+ \frac{1}{\frac{1 - x}{\sqrt{1 + x^{2}}}} + \sqrt{1 + \frac{{(1 - x)}^{2}}{1 + x^{2}}} .

\int_{α (x)}^{β (x)} c h (u) s h (u) d u = \frac{1}{2} \int_{α (x)}^{β (x)} s h (2 u) d u = \frac{1}{4} {[c h (2 u)]}_{α (x)}^{β (x)}

= \frac{1}{8} {[e^{2 u} - e^{- 2 u}]}_{α (x)}^{β (x)} = \frac{1}{8} [e^{2 β (x)} - e^{- 2 β (x)} - e^{2 α (x)} + e^{- 2 α (x)}]

w i t α (x) = a r g c h (\frac{1 - x}{\sqrt{1 + x^{2}}}) a n d β (x) = a r g c h (\frac{2 - x}{\sqrt{1 + x^{2}}})

t h e v a l u e o f f (x) i s k n o w n \dots

Commented by mathmax by abdo last updated on 23/Aug/19

2) w e h a v e f^{^{'}} (x) = \int_{1}^{2} (t + 1) \frac{(- 2 t)}{2 \sqrt{t^{2} - 2 x t - 1}} d t

= - 2 \int_{1}^{2} \frac{t^{2} + t}{\sqrt{t^{2} - 2 x t - 1}} d t \Rightarrow \int_{1}^{2} \frac{t^{2} + t}{\sqrt{t^{2} - 2 x t - 1}} d t = - \frac{1}{2} f^{^{'}} (x)

r e s t t o c a l c u l a t e f^{^{'}} (x) \dots

Commented by mathmax by abdo last updated on 23/Aug/19

3) l e t I = \int_{1}^{2} (t + 1) \sqrt{t^{2} - t - 1} d t w e h a v e t^{2} - t - 1 =

t^{2} - 2 \frac{t}{2} + \frac{1}{4} - 1 - \frac{1}{4} = {(t - \frac{1}{2})}^{2} - \frac{5}{4} w e d o t h e c h a n g e m e n t

t - \frac{1}{2} = \frac{\sqrt{5}}{2} c h (u) \Rightarrow u = a r g c h (\frac{2 t - 1}{\sqrt{5}}) \Rightarrow

I = \int_{a r g c h (\frac{1}{\sqrt{5}})}^{a r g c h (\frac{3}{\sqrt{5}})} (\frac{1}{2} + \frac{\sqrt{5}}{2} c h (u)) \frac{\sqrt{5}}{2} s h (u) d u

= \frac{\sqrt{5}}{4} \int_{l n (\frac{1}{\sqrt{5}} + \sqrt{1 + \frac{1}{5}})}^{l n (\frac{3}{\sqrt{5}} + \sqrt{1 + \frac{9}{5}})} s h (u) d u + \frac{5}{8} \int_{l n (\frac{1}{\sqrt{5}} + \sqrt{1 + \frac{1}{5}})}^{l n (\frac{3}{\sqrt{5}} + \sqrt{1 + \frac{9}{5}})} s h (2 u) d u

= \frac{\sqrt{5}}{8} {[e^{u} - e^{- u}]}_{l n (\frac{1}{\sqrt{5}} + \frac{\sqrt{6}}{\sqrt{5}})}^{l n (\frac{3}{\sqrt{5}} + \frac{\sqrt{13}}{\sqrt{5}})} + \frac{5}{16} {[e^{2 u} - e^{- 2 u}]}_{l n (\frac{1 + \sqrt{6}}{\sqrt{5}})}^{l n (\frac{3 + \sqrt{13}}{\sqrt{5}})}

= \frac{\sqrt{5}}{8} {\frac{3 + \sqrt{13}}{\sqrt{5}} - \frac{1}{\frac{3 + \sqrt{13}}{\sqrt{5}}} - (\frac{1 + \sqrt{6}}{\sqrt{5}}) + \frac{1}{\frac{1 + \sqrt{6}}{\sqrt{5}}}}

+ \frac{5}{16} {{(\frac{3 + \sqrt{13}}{\sqrt{5}})}^{2} - \frac{1}{{(\frac{3 + \sqrt{13}}{\sqrt{5}})}^{2}} - {(\frac{1 + \sqrt{6}}{\sqrt{5}})}^{2} + \frac{1}{{(\frac{1 + \sqrt{6}}{\sqrt{5}})}^{2}}}