Integrate log(sinx) from 0 to pi /2?

1 Answer
Sep 26, 2017

Answer:

#I=int_0^(pi/2)logsinxdx=-(pi/2)log2#

Explanation:

We use the property #int_0^af(x)dx=int_0^af(a-x)dx#

hence we can write #I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logsin(pi/2-x)dx#

or #I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logcosxdx#

or #2I=int_0^(pi/2)(logsinx+logcosx)dx=int_0^(pi/2)log(sinxcosx)dx#

= #int_0^(pi/2)log((sin2x)/2)dx=int_0^(pi/2)(logsin2x-log2)dx#

= #int_0^(pi/2)logsin2xdx-int_0^(pi/2)log2dx#

= #int_0^(pi/2)logsin2xdx-(pi/2)log2# .............(A)

Let #I_1=int_0^(pi/2)logsin2xdx# and #t=2x#, then #I_1=1/2int_0^pilogsintdt#

and using the property #int_0^(2a)f(x)dx=2int_0^af(a-x)dx#, if #f(2a-x)=f(x)# - note that here #logsint=logsin(pi-t)# and we get

#I_1=1/2int_0^pilogsintdt=int_0^(pi/2)logsintdt=I#

Hence (A) becomes #2I=I-(pi/2)log2#

or #I=-(pi/2)log2#