find-the-value-of-1-1-dx-1-x-2-1-x-2-

Question Number 27215 by abdo imad last updated on 03/Jan/18

f i n d t h e v a l u e o f \int_{- 1}^{1} \frac{d x}{\sqrt{1 - x^{2}} + \sqrt{1 + x^{2}}} .

Answered by Giannibo last updated on 03/Jan/18

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

\underset{- 1}{\int^{1}} \frac{\sqrt{1 - x^{2}} - \sqrt{1 + x^{2}}}{2} dx

= \lim_{u \to 1} (\underset{- u}{\int^{u}} \frac{\sqrt{1 - x^{2}}}{2} dx - \underset{- u}{\int^{u}} \frac{\sqrt{1 + x^{2}}}{2} dx)

= \lim_{u \to 1} \frac{1}{4} (\sin^{- 1} (u) + u \sqrt{1 - u^{2}} - \sin^{- 1} (- u) + u \sqrt{1 - u^{2}} - (\ln (∣ \sqrt{u^{2} + 1} + u ∣) + u \sqrt{u^{2} + 1} - \ln (∣ \sqrt{u^{2} + 1} - u ∣) + u \sqrt{u^{2} + 1})

= \frac{1}{4} (\frac{π}{2} + 0 + \frac{π}{2} - (\ln (\sqrt{2} + 1) + \sqrt{2} - \ln (\sqrt{2} - 1) + \sqrt{2}))

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

Commented by abdo imad last updated on 05/Jan/18

\int_{- 1}^{1} \frac{d x}{\sqrt{1 - x^{2}} + \sqrt{1 + x^{2}}} = \int_{- 1}^{1} \frac{\sqrt{1 - x^{2}} - \sqrt{1 + x^{2}}}{- 2 x^{2}} d x \dots!!