please-prove-that-0-pi-2-log-sinx-dx-pi-2-log2-or-0-pi-2-log-cosx-dx-pi-2-log2-

Question Number 24852 by nnnavendu last updated on 27/Nov/17

please prove that \int_{0}^{\frac{π}{2}} \log (sinx) dx = - \frac{π}{2} \log 2

or

\int_{0}^{\frac{π}{2}} \log (cosx) dx = - \frac{π}{2} \log 2

Commented by Tinku Tara last updated on 28/Nov/17

\sin (x) = \cos (\frac{π}{2} - x)

\Rightarrow \int_{0}^{π / 2} \log (\cos x)

substitute

u = \frac{π}{2} - x \Rightarrow x = 0, u = \frac{π}{2}, x = \frac{π}{2}, u = 0

d u = - d x

\int_{0}^{π / 2} \log (\cos x) = - \int_{π / 2}^{0} \log (\sin u) d u

= \int_{0}^{π / 2} \log (\sin u) d u

\int_{0}^{π / 2} \log (\sin x) d x = \int_{0}^{π / 2} \log (\cos x) d x = I

2 I = \int_{0}^{π / 2} [\log (\sin x) + \log (\cos x)] d x

= \int_{0}^{π / 2} \log \frac{\sin 2 x}{2} d x

= \int_{0}^{π / 2} \log (\sin 2 x) d x - \int_{0}^{π / 2} \ln 2 d x

\int_{0}^{π / 2} \log (\sin 2 x) d x

2 x = u \Rightarrow d x = d u / 2

l i m i t s change to 0 to π

= \frac{1}{2} \int_{0}^{π} \log (\sin u) d u - \frac{π}{2} \ln 2

= \frac{1}{2} [\int_{0}^{π / 2} \log (\sin u) d u + \int_{π / 2}^{π} \log (\sin u) d u] - \frac{π}{2} \ln 2

= \frac{1}{2} [I + \int_{π / 2}^{π} \log (\sin u) d u] - \frac{π}{2} \ln 2

t = u - \frac{π}{2} \Rightarrow u = \frac{π}{2} + t \Rightarrow \sin u = \cos t

d t = d u, limits change to 0 to \frac{π}{2}

= \frac{1}{2} [I + \int_{0}^{π / 2} \log (\cos t) d t] - \frac{π}{2} \ln 2

2 I = \frac{1}{2} [I + I] - \frac{π}{2} \ln 2

I = - \frac{π}{2} \ln 2