Evaluate-x-2-a-2-dx-

Question Number 12098 by Nayon last updated on 13/Apr/17

E v a l u a t e \int \sqrt{x^{2} - a^{2}} d x

Answered by sma3l2996 last updated on 13/Apr/17

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

l e t u = \frac{x}{a} \Rightarrow d u = \frac{d x}{a}

A = a^{2} \int \sqrt{u^{2} - 1} d u

l e t u = c o s h (t) \Rightarrow d u = s i n h (t) d t

s i n h (t) d u = {s i n h}^{2} t d t \Leftrightarrow \sqrt{{c o s h}^{2} t - 1} d u = {s i n h}^{2} t d t

\sqrt{u^{2} - 1} d u = {s i n h}^{2} t d t

A = a^{2} \int {s i n h}^{2} t d t = a^{2} \int \frac{c o s h (2 t) - 1}{2} d t

A = \frac{a^{2}}{2} (\frac{s i n h (2 t)}{2} - t) + C

A = \frac{a^{2}}{4} (s i n h (2 a c o s h (\frac{x}{a})) - a c o s h (\frac{x}{a})) + C

Commented by Nayon last updated on 13/Apr/17

p l s a n s w i t h o u t u s i n g h y p e r b o l i c

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Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

x = a i t, d x = a i d t, i = \sqrt{- 1}

I = \int \sqrt{- a^{2} t^{2} - a^{2}} a i d t = - a^{2} \int \sqrt{t^{2} + 1} d t =

= - a^{2} [t \sqrt{t^{2} + 1} - \int t \frac{2 t d t}{2 \sqrt{t^{2} + 1}}] =

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

= - a^{2} [t \sqrt{t^{2} + 1} - \int \sqrt{t^{2} + 1} d t + \int \frac{d t}{\sqrt{t^{2} + 1}}] =

- a^{2} t \sqrt{t^{2} + 1} + a^{2} \int \sqrt{t^{2} + 1} d t - \frac{a^{2}}{2} l n (t + \sqrt{t^{2} + 1})

2 I = - a^{2} \frac{x}{a i} . \frac{\sqrt{x^{2} - a^{2}}}{a i} - \frac{a^{2}}{2} l n (\frac{x}{a i} + \frac{\sqrt{x^{2} - a^{2}}}{a i}) \Rightarrow

I = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{4} l n (x + \sqrt{x^{2} - a^{2}}) + C .

(- a^{2} \times \frac{1}{a i} \times \frac{1}{a i} = 1, C = \frac{a^{2}}{4} l n (a i))