show that \int_{0}^{\pi /2}{sin\theta logsin\theta }d\theta = log\frac{2}{e}
1MERA LAGBHAG HO GAYA.
CONVERT sin \theta =2sin\frac{\theta }{2}cos\frac{\theta }{2} both inside log and outside log...
then use property log xy = log x + log y and expand.
u will get integral of the form ∫ t log t dt integrate it by parts and try to find the answer...
29The Integration can be written as
\int 2sin\frac{\theta }{2}cos\frac{\theta }{2} log(2sin\frac{\theta }{2}cos\frac{\theta }{2}) = \int 2sin\frac{\theta }{2}cos\frac{\theta }{2} log2sin\frac{\theta }{2} +\int 2sin\frac{\theta }{2}cos\frac{\theta }{2}logcos\frac{\theta }{2}
2sin\frac{\theta }{2} = t ....and... cos\frac{\theta }{2} = z
\frac{dt}{d\theta } = cos\frac{\theta }{2} ...and...\frac{2dz}{d\theta } = -sin\frac{\theta }{2}
we know that \int tlogtdt = \frac{t^{2}}{2}logt - \frac{t^{2}}{4}...
now the limit for t is from 0 to √2 and for z is from 1 to 1/√2...
I hope u can finish it off from here
29ok Manmay now wat abt this one..
\int_{0}^{\pi /2}{cos\theta logcos\theta d\theta }
what substitution will u make in this case?
1PROBABLY cos\theta =t
cos\theta =t \Rightarrow -sin \theta d\theta = dt \Rightarrow d\theta =-\frac{1}{\sqrt{1-t^{2}}}
which gives, \int_{0}^{1}{} \frac{logt}{\sqrt{1-t^{2}}}dt
now may be by parts will do..
29@Manmay..actually a very easy solution exist..Now that the first question is solved...think abt some Definite Integrals Properties...