calculate-1-1-1-x-2-1-x-2-dx-

Question Number 63509 by mathmax by abdo last updated on 05/Jul/19

c a l c u l a t e \int_{- 1}^{1} (\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}) d x

Commented by Prithwish sen last updated on 05/Jul/19

2 \int_{0}^{1} (\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}) dx

= 2 {[\frac{x}{2} \sqrt{1 + x^{2}} + \frac{1}{2} \ln ∣ x + \sqrt{1 + x^{2}} ∣ - \frac{x}{2} \sqrt{1 - x^{2}} - \frac{1}{2} \sin^{- 1} x]}_{0}^{1}

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

please check .

Answered by MJS last updated on 05/Jul/19

- \underset{- 1}{\int^{1}} \sqrt{1 - x^{2}} d x = - \frac{π}{2}

\underset{- 1}{\int^{1}} \sqrt{1 + x^{2}} d x = \frac{1}{2} {[x \sqrt{1 + x^{2}} + \ln (x + \sqrt{1 + x^{2}})]}_{- 1}^{1} = \sqrt{2} + \ln (1 + \sqrt{2})

\underset{- 1}{\int^{1}} (\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}) d x = \sqrt{2} + \ln (1 + \sqrt{2}) - \frac{π}{2}

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