0-pi-2-log-sin-d- – Tinku Tara

Question Number 313 by Vishal Bhardwaj last updated on 25/Jan/15

\int_{0}^{\frac{π}{2}} l o g s i n θ d θ

Answered by prakash jain last updated on 20/Dec/14

I = \int_{0}^{π / 2} \ln \sin θ d θ \dots . . (i)

put θ = \frac{π}{2} - α, d θ = - d α

I = \int_{π / 2}^{0} \ln \sin (\frac{π}{2} - α) (- d α)

= - \int_{π / 2}^{0} \ln \cos α d α = \int_{0}^{π / 2} \ln \cos α d α

= \int_{0}^{π / 2} \ln \cos θ d θ \dots . (i i)

add (i) and (i i)

2 I = \int_{0}^{π / 2} (\ln \sin θ + \ln \cos θ) d θ

2 I = \int_{0}^{π / 2} \ln \frac{\sin 2 θ}{2} d θ

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

2 I = \int_{0}^{π / 2} \ln \sin 2 θ d θ - \int_{0}^{π / 2} \ln 2 d θ

In first integral put 2 θ = β, 2 d θ = d β so

the equation one now with updated limits

2 I = \int_{0}^{π} \ln \sin β \frac{d β}{2} - \int_{0}^{π / 2} \ln 2 d θ

2 I = \frac{1}{2} \int_{0}^{π} \ln \sin β d β - \int_{0}^{π / 2} \ln 2 d θ \dots . (i i i)

Now we will evaluate first integral

\int_{0}^{π} \ln \sin β d β = \int_{0}^{π / 2} \ln \sin β d β + \int_{π / 2}^{π} \ln \sin β d β

\int_{0}^{π} \ln \sin β d β == I + \int_{π / 2}^{π} \ln \sin β d β

put β = π - t d β = - d t

\int_{0}^{π} \ln \sin β d β == I + \int_{π / 2}^{0} \ln \sin (π - t) (- d t)

\int_{0}^{π} \ln \sin β d β == I - \int_{π / 2}^{0} \ln \sin t d t

\int_{0}^{π} \ln \sin β d β == I + \int_{0}^{π / 2} \ln \sin t d t

\int_{0}^{π} \ln \sin β d β == I + I = 2 I \dots . (i v)

putting this value in (i i i)

2 I = \frac{1}{2} (2 I) - \int_{0}^{π / 2} \ln 2 d θ

2 I = I - \frac{π}{2} \ln 2

I = - \frac{π}{2} \ln 2