find-g-a-x-a-x-2-a-2-dx-

Question Number 62209 by maxmathsup by imad last updated on 17/Jun/19

f i n d g (a) = \int (x + a) \sqrt{x^{2} - a^{2}} d x

Commented by maxmathsup by imad last updated on 18/Jun/19

c h a n g e m e n t x = a c h (t) g i v e g (a) = \int (a c h t + a) \sqrt{a^{2} ({c h}^{2} t - 1)} a s h (t) d t

= a^{2} ∣ a ∣ \int (1 + c h (t) {s h}^{2} (t) d t

= a^{2} ∣ a ∣ \int 2 {c h}^{2} t {s h}^{2} t d t = 2 a^{2} ∣ a ∣ \int {(\frac{1}{2} s h (2 t))}^{2} d t

= \frac{a^{2} ∣ a ∣}{2} \int \frac{c h (4 t) - 1}{2} d t = \frac{a^{2} ∣ a ∣}{4} \int (c h (4 t) - 1) d t

= \frac{a^{2} ∣ a ∣}{16} s h (4 t) - \frac{a^{2} ∣ a ∣}{4} t + c b u t w e h a v e

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

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

s h (4 t) = \frac{e^{4 t} - e^{- 4 t}}{2} = \frac{1}{2} {{(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1})}^{4} - {(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1})}^{- 4}} \Rightarrow

A = \frac{a^{2} ∣ a ∣}{32} {{(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1})}^{4} - {(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}}} - 1)}^{- 4}} - \frac{a^{2} ∣ a ∣}{4} l n (\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1}) + C

Answered by tanmay last updated on 17/Jun/19

\int x \sqrt{x^{2} - a^{2}} d x + a \int \sqrt{x^{2} - a^{2}} d x

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

\frac{1}{2} \times \frac{{(x^{2} - a^{2})}^{\frac{3}{2}}}{\frac{3}{2}} + a [\frac{x \sqrt{x^{2} - a^{2}}}{2} - \frac{a^{2}}{2} l n (x + \sqrt{x^{2} - a^{2}})] + c

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