how-do-we-find-0-pi-2-sinh-1-x-dx-and-0-pi-2-cosh-1-xdx-

Question Number 76232 by Rio Michael last updated on 25/Dec/19

h o w d o w e f i n d

\underset{0}{\int^{\frac{π}{2}}} {s i n h}^{- 1} x d x a n d \underset{0}{\int^{\frac{π}{2}}} \cosh^{- 1} x d x

Commented by mr W last updated on 25/Dec/19

\cosh^{- 1} x i s d e f i n e d f o r x ⩾ 1!

Commented by turbo msup by abdo last updated on 25/Dec/19

l e t I = \int_{0}^{\frac{π}{2}} a r g s h (x) d x

\Rightarrow I =_{a r g s h (x) = t} \int_{0}^{a r g s h (\frac{π}{2})} t c h (t) d t

= \int_{0}^{l n (\frac{π}{2} + \sqrt{1 + \frac{π^{2}}{4}})} t c h t () d t

=_{b y p a r t s} {[t s h (t)]}_{0}^{l n (\frac{π}{2} + \sqrt{1 + \frac{π^{2}}{4}})}

- \int_{0}^{l n (\frac{π}{2} + \sqrt{1 + \frac{π^{2}}{4}})} s h (t) d t

= l n (\frac{π}{2} + \sqrt{1 + \frac{π^{2}}{4}}) \frac{\frac{π}{2} + \sqrt{1 + \frac{π^{2}}{4}} - {(\frac{π}{2} + \sqrt{1 + \frac{π^{2}}{4}})}^{- 1}}{2}

- \frac{1}{2} (\frac{π}{2} + \sqrt{1 + \frac{π^{2}}{4}}) + {(\frac{π}{2} + \sqrt{1 + \frac{π^{2}}{4}})}^{- 1})

Answered by mind is power last updated on 25/Dec/19

u = {sh}^{-} (x)

by part

\int {sh}^{-} (x) dx = [{xsh}^{-} (x)] - \int \frac{x}{\sqrt{1 + x^{2}}} dx

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

sam with {ch}^{-} (x)

Answered by mr W last updated on 25/Dec/19

\int_{0}^{\frac{π}{2}} \sinh^{- 1} x d x

= \frac{π}{2} \sinh^{- 1} (\frac{π}{2}) - \int_{0}^{\sinh^{- 1} (\frac{π}{2})} \sinh y d y

= \frac{π}{2} \sinh^{- 1} (\frac{π}{2}) - [\cosh (\sinh^{- 1} \frac{π}{2}) - 1]

= \frac{π}{2} \ln (\frac{π}{2} + \sqrt{1 + {(\frac{π}{2})}^{2}}) + 1 - \sqrt{1 + {(\frac{π}{2})}^{2}}

\approx 1.075329

Commented by mr W last updated on 25/Dec/19

Commented by mr W last updated on 25/Dec/19

I_{1} = \int_{0}^{\frac{π}{2}} \sinh^{- 1} x d x

I_{2} = \int_{0}^{\sinh^{- 1} \frac{π}{2}} \sinh y d y

I_{1} + I_{2} = \frac{π}{2} \times \sinh^{- 1} \frac{π}{2}

I_{2} i s e a s i e r t o g e t t h a n I_{1} .

Commented by Rio Michael last updated on 25/Dec/19

t h a n k s

Commented by mr W last updated on 25/Dec/19

You can't use 'macro parameter character #' in math mode

Facebook Tweet Pin