Prove that \(\int_{0}^{\pi/2}\) log(sinx)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx = -\(\pi\over 2\)log 2.

Solution :

Let I = \(\int_{0}^{\pi/2}\) log(sinx)dx    …….(i)

then I = \(\int_{0}^{\pi/2}\) \(log(sin({\pi\over 2}-x))\)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx     …….(ii)

Adding (i) and (ii), we get

2I = \(\int_{0}^{\pi/2}\) log(sinx)dx + \(\int_{0}^{\pi/2}\) log(cosx)dx = \(\int_{0}^{\pi/2}\) (log(sinx)dx + log(cosx))

\(\implies\) \(\int_{0}^{\pi/2}\) log(sinxcosx)dx = \(\int_{0}^{\pi/2}\) \(log({2sinxcosx\over 2})\)dx

= \(\int_{0}^{\pi/2}\) \(log({sin2x\over 2})\)dx = \(\int_{0}^{\pi/2}\) log(sin2x)dx – \(\int_{0}^{\pi/2}\) log(2)dx

= \(\int_{0}^{\pi/2}\) log(sin2x)dx – (log 2)\({(x)}^{\pi/2}_{0}\)

\(\implies\)  2I = \(\int_{0}^{\pi/2}\) log(sin2x)dx – \(\pi\over 2\)log 2   …(iii)

Let \(I_1\) = \(\int_{0}^{\pi/2}\) log(sin2x)dx,  putting 2x = t, we get

\(I_1\) = \(\int_{0}^{\pi}\) log(sint)\(dt\over 2\) = \(1\over 2\) \(\int_{0}^{\pi}\) log(sint)dt = \(1\over 2\) 2\(\int_{0}^{\pi/2}\) log(sint)dt

\(I_1\) = \(\int_{0}^{\pi/2}\) log(sinx)dx

\(\therefore\)  (iii) becomes; 2I = I – \(\pi\over 2\)log 2

Hence  \(\int_{0}^{\pi/2}\) log(sinx)dx = – \(\pi\over 2\)log 2


Similar Questions

What is the integration of x tan inverse x dx ?

What is the integration of tan inverse root x ?

Evaluate : \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\)

Evaluate : \(\int\) \(dx\over {3sinx + 4cosx}\)

Leave a Comment

Your email address will not be published.