# ( integral )

• May 15th 2009, 03:09 PM
a.h.m.a.d
( integral )
• May 15th 2009, 04:12 PM
NonCommAlg
Quote:

Originally Posted by a.h.m.a.d

the integrand is an even function. so the integral is equal to $I=2 \int_0^1 \ln(\sqrt{1-x} + \sqrt{1+x}) \ dx.$ now by parts gives us:

$I=\ln 2 + \int_0^1 \frac{x^2 \ dx}{\sqrt{1-x^2}(1 + \sqrt{1-x^2})}.$ finally put $x = \sin \theta,$ to get: $I=\ln 2 + \int_0^{\frac{\pi}{2}} (1 - \cos \theta) \ d \theta=\ln 2 + \frac{\pi}{2} - 1.$
• May 15th 2009, 06:13 PM
Krizalid
$\int_{0}^{1}{\ln \left( \sqrt{1+x}+\sqrt{1-x} \right)\,dx}=\int_{0}^{1}{\ln \left( \sqrt{1+x}\left( 1+\sqrt{\frac{1-x}{1+x}} \right) \right)\,dx},$ besides $\frac{1}{2}\int_{0}^{1}{\ln (1+x)\,dx}$ and $\int_{0}^{1}{\ln \left( 1+\sqrt{\frac{1-x}{1+x}} \right)\,dx}$ are pretty straightforward to compute.