If n is any positive integer, prove that integral : #int_0^pisen^2(nx)dx=pi/2# ?

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#int_0^pisen^2(nx)dx=pi/2#

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Answer 1 · 2018-04-10T20:02:31

Please see below.

.

#int_0^pisin^2(nx)dx#

#sin^2(nx)=(1-cos(2nx))/2#

#int_0^pi(1-cos(2nx))/2dx=1/2int_0^pidx-1/2int_0^picos(2nx)dx=(1/2x-1/(2n)sin(2nx))_0^pi=pi/2-1/(2n)sin(2npi)-0+0=pi/2-sin(2npi)/(2n)#

#n=("any integer"), :. sin(2npi)=0#

#int_0^pisin^2(nx)dx=pi/2#

Answer 2 · 2018-04-10T20:15:55

See below.

#int_0^pi"sen"^2(nx)dx#

Using the double angle theorem:

#=int_0^(pi) 1/2(1-cos(2nx))dx#

Pull out #1/2#:

#=1/2 int_0^(pi) (1-cos(2nx))dx#

Integral of difference is the difference of integrals:

#=1/2 int_0^(pi) dx - int_0^{pi}cos(2nx)dx#

Evaluating the integrals:

#=1/2[pi - 0] - 1/(2n)["sen"(2nx)]_{0}^{pi}#

#=pi/2 - 1/(2n)[0 - 0]#

#= pi/2#

Hence:

#=>int_0^pi"sen"^2(nx)dx = pi/2#