What is #f(x) = int cot2x dx# if #f(pi/8) = 0 #?

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Answer 1 · 2016-11-29T16:36:48

#f(x)=1/2lnsqrt2sin(2x)#

#f(x)=intcot2xdx#

#f(x)=int((cos2x)/(sin2x))dx#

now#""(d/(dx)(sin2x)=2cos2x)#

we have a log integration.

#f(x)=int((cos2x)/(sin2x))dx=1/2lnsin2x+C#
#f(pi/8)=0#

#:.1/2lnsin(pi/4)+C=0#

rewrite #C=1/2lnK" "#in order to include it in the log.

#1/2lnsin(pi/4)+1/2lnk=0#

#1/2[(lnsin(pi/4)+lnk]=0#

#1/2(lnksin(pi/4))=0#

#lnx=0=>x=1#

#:. ksin(pi/4)=1#

#k(sqrt2/2)=1#

#k=2/sqrt2=sqrt2#

#f(x)=1/2lnsqrt2sin(2x)#