solve-a-1-2-a-2-cosh-1-1-a-x-dx-with-a-gt-0-

Question Number 122631 by MJS_new last updated on 18/Nov/20

solve \underset{{(a - 1)}^{2}}{\int^{a^{2}}} \cosh^{- 1} \frac{1}{\sqrt{a - \sqrt{x}}} d x with a > 0

Commented by liberty last updated on 18/Nov/20

w a w . . n i c e q u e s t i o n

Commented by MJS_new last updated on 19/Nov/20

f (a) = \underset{{(a - 1)}^{2}}{\int^{a^{2}}} \cosh^{- 1} \frac{1}{\sqrt{a - \sqrt{x}}} d x is linear for a ⩾ 1

I found a nice path to solve it, will post it later

Answered by mathmax by abdo last updated on 18/Nov/20

I = \int_{{(a - 1)}^{2}}^{a^{2}} argch (\frac{1}{\sqrt{a - \sqrt{x}}}) dx chamgement \sqrt{a - \sqrt{x}} = t give

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

I = \int_{1}^{o} argch (\frac{1}{t}) 4 t (t^{2} - a) dt = - 4 \int_{o}^{1} t (t^{2} - a) \ln (\frac{1}{t} + \sqrt{\frac{1}{t^{2}} - 1}) dt

= - 4 \int_{0}^{1} t (t^{2} - a) \ln (\frac{1 + \sqrt{1 - t^{2}}}{t}) dt

=_{t = \sin θ} - 4 \int_{0}^{\frac{π}{2}} \sin θ (\sin^{2} θ - a) \ln (\frac{1 + \cos θ}{\sin θ}) \cos θ d θ

= - 4 \int_{0}^{\frac{π}{2}} \sin θ (\sin^{2} θ - a) \ln (\frac{{2 \cos}^{2} (\frac{θ}{2})}{2 \cos (\frac{θ}{2}) \sin (\frac{θ}{2})}) \cos θ d θ

= 4 \int_{0}^{\frac{π}{2}} \sin θ \cos θ (\sin^{2} θ - a) \ln (\tan (\frac{θ}{2})) d θ

= 4 \int_{0}^{\frac{π}{2}} (\sin^{3} θ \cos θ - \sin θ \cos θ) \ln (\tan (\frac{θ}{2})) d θ this integral can be

solved by parts if we put u^{^{'}} = \sin^{3} \cos θ - \sin θ \cos θ and v = \ln (\tan (\frac{θ}{2}))

\dots be continued \dots

Commented by MJS_new last updated on 19/Nov/20

thank you for trying

Answered by MJS_new last updated on 19/Nov/20

\int \cosh^{- 1} \frac{1}{\sqrt{a - \sqrt{x}}} d x =

by parts but with a simple trick :

u^{^{'}} = 1 \to u = x - a^{2}

v = \cosh^{- 1} \frac{1}{\sqrt{a - \sqrt{x}}} \to v^{'} = - \frac{1}{4 \sqrt{x} (\sqrt{x} - a) \sqrt{\sqrt{x} - a + 1}}

= (x - a^{2}) \cosh^{- 1} \frac{1}{\sqrt{a - \sqrt{x}}} + \frac{1}{4} \int \frac{x - a^{2}}{\sqrt{x} (\sqrt{x} - a) \sqrt{\sqrt{x} - a + 1}} d x =

\frac{1}{4} \int \frac{x - a^{2}}{\sqrt{x} (\sqrt{x} - a) \sqrt{\sqrt{x} - a + 1}} d x =

[t = \sqrt{\sqrt{x} - a + 1} \to d x = 4 \sqrt{x} \sqrt{\sqrt{x} - a + 1} d t]

= \int (t^{2} + 2 a - 1) d t = \frac{1}{3} t (t^{2} + 3 (2 a - 1)) =

= \frac{1}{3} (\sqrt{x} + 5 a - 2) \sqrt{\sqrt{x} - a + 1}

= (x - a^{2}) \cosh^{- 1} \frac{1}{\sqrt{a - \sqrt{x}}} + \frac{1}{3} (\sqrt{x} + 5 a - 2) \sqrt{\sqrt{x} - a + 1} + C

inserting the borders we get

f (a) = (2 a - 1) \cosh^{- 1} \frac{1}{\sqrt{a - ∣ a - 1 ∣}} - \frac{(5 a + ∣ a - 1 ∣ - 2) \sqrt{1 - a + ∣ a - 1 ∣}}{3} + \frac{(5 a + ∣ a ∣ - 2) \sqrt{1 - a + ∣ a ∣}}{3}

for 0 < a < 1 we get

f (a) = (2 a - 1) \cosh^{- 1} \frac{1}{\sqrt{2 a - 1}} - \frac{(4 a - 1) \sqrt{2 (1 - a)}}{3} + \frac{2 (3 a - 1)}{3}

for a ⩾ 1 we get

f (a) = \frac{2 (3 a - 1)}{3}

Commented by liberty last updated on 19/Nov/20

b y p a r t s \int v d u = v u - \int u d v ?

Commented by MJS_new last updated on 19/Nov/20

\int u^{'} v = u v - \int {u v}^{'}

Commented by liberty last updated on 19/Nov/20

h a h a . . {i t}^{'} s s a m e s i r

Commented by MJS_new last updated on 19/Nov/20

yes

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