∫_0 ^(π/2) ln(sinx+cosx)dx=? −−−−−−−−−−−−by M.A


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Question Number 166260 by amin96 last updated on 16/Feb/22

$\int_{0}^{\frac{π}{2}} \ln (sinx + cosx) dx = ?$ $- - - - - - - - - - - - by M . A$
	Answered by Eulerian last updated on 17/Feb/22

	$I = \int_{0}^{\frac{π}{2}} \ln (\sin x + \cos x) dx = \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (1 + \sin 2 x) dx$ $By substitution, let :$ $z = 2 x$ $\frac{1}{2} dz = dx$ $∴$ $I = \frac{1}{4} \int_{0}^{π} \ln (1 + \sin z) dz = \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (1 + \sin z) dz$ $By noticing that it somehow looks like the integral representation of {Catalan}^{'} s$ $constant, such that :$ $G = \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\sec z + \tan z) dz = \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (1 + \sin z) - \ln (\cos z) dz$ $∴$ $I = \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (1 + \sin z) - \ln (\cos z) + \ln (\cos z) dz$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (1 + \sin z) - \ln (\cos z) dz + \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\cos z) dz$ $= G + \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz (King Rule)$ $By adding those integrals, we get :$ $2 I = 2 G + \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz + \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\cos z) dz$ $I = G + \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\sin 2 z) - \ln (2) dz$ $= G + \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\sin 2 z) dz - \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (2) dz$ $= G + \frac{1}{8} \int_{0}^{π} \ln (\sin z) dz - \frac{π \ln (2)}{8}$ $= G - \frac{π \ln (2)}{8} + \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz$ $From what we have earlier, by comparing to what we have right now, we get :$ $G + \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz = G - \frac{π \ln (2)}{8} + \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz$ $\frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz = - \frac{π \ln (2)}{8} + \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz$ $\frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz - \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\sin z) dz = - \frac{π \ln (2)}{8}$ $\int_{0}^{\frac{π}{2}} \ln (\sin z) dz = - \frac{π \ln (2)}{2}$ $Finally, we now have our answer to integral I :$ $I = G - \frac{π \ln (2)}{8} - (\frac{1}{4}) \cdot \frac{π \ln (2)}{2} = G - \frac{π \ln (2)}{4}$ $Solution by : Kevin$
	Commented by amin96 last updated on 17/Feb/22

	$correct answer bravo sir Kevin$
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