Find-0-2-ln-sinx-1-sin-2-x-dx-

Question Number 205432 by hardmath last updated on 21/Mar/24

Find : Ω = \int_{0}^{2 π} \ln (sinx + \sqrt{1 + \sin^{2} x}) dx

Answered by MathedUp last updated on 21/Mar/24

0

cauz - f (z) = f (- z), f (z + π) = - f (z), f (z + 2 π) = f (z)

So, f (z) is periodic function that having period π

\int_{0}^{2 π} f (z) d z = 0

Answered by mr W last updated on 21/Mar/24

Ω = \int_{0}^{2 π} \ln (\sin x + \sqrt{1 + \sin^{2} x}) d x

= \int_{0}^{π} \ln (\sin x + \sqrt{1 + \sin^{2} x}) d x + \int_{π}^{2 π} \ln (\sin x + \sqrt{1 + \sin^{2} x}) d x

= \int_{0}^{π} \ln (\sin x + \sqrt{1 + \sin^{2} x}) d x + \int_{0}^{π} \ln (\sin (x + π) + \sqrt{1 + \sin^{2} (x + π)}) d (x + π)

= \int_{0}^{π} \ln (\sin x + \sqrt{1 + \sin^{2} x}) d x + \int_{0}^{π} \ln (- \sin x + \sqrt{1 + \sin^{2} x}) d x

= \int_{0}^{π} \ln (\sin x + \sqrt{1 + \sin^{2} x}) d x + \int_{0}^{π} \ln \frac{1}{\sin x + \sqrt{1 + \sin^{2} x}} d x

= \int_{0}^{π} \ln (\sin x + \sqrt{1 + \sin^{2} x}) d x - \int_{0}^{π} \ln (\sin x + \sqrt{1 + \sin^{2} x}) d x

= 0

Commented by hardmath last updated on 21/Mar/24

cool dear professor thankyou