What is int_(pi/2)^pi lnsinx?

1 Answer
Dec 4, 2016

${\int}_{\frac{\pi}{2}}^{\pi} \ln \left(\sin x\right) \mathrm{dx} = \frac{\pi}{2} \ln \left(\frac{1}{2}\right)$

Explanation:

Here you can see how to calculate that:

${\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin x\right) \mathrm{dx} = \frac{\pi}{2} \ln \left(\frac{1}{2}\right)$

To calculate:

${\int}_{\frac{\pi}{2}}^{\pi} \ln \left(\sin x\right) \mathrm{dx}$

just substitute:

$t = \pi - x$

and you have:

${\int}_{\frac{\pi}{2}}^{\pi} \ln \left(\sin x\right) \mathrm{dx} = {\int}_{\frac{\pi}{2}}^{0} \ln \left(\sin \left(\pi - t\right)\right) \left(- \mathrm{dt}\right) = {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin t\right) \mathrm{dt}$

because:

$\sin \left(\pi - t\right) = \sin t$

and for the properties of the definite integral:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = - {\int}_{b}^{a} f \left(x\right) \mathrm{dx}$