What is the integral of #int sin^2(πx / 2)#?

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Answer 1 · 2017-03-14T03:00:13

#1/2x-1/(2pi)sin(pix)+C#

Use the form of the cosine double-angle identity with sine in it:

#cos(2alpha)=1-2sin^2(alpha)#

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

Which implies that:

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

Then:

#intsin^2((pix)/2)dx=1/2int(1-cos(pix))dx#

Integrating both of these, and integrating #cos(pix)# with substitution:

#=1/2intdx-1/2intcos(pix)dx#

Let #u=pix# to #du=pidx#:

#=1/2x-1/(2pi)intcos(pix)(pidx)=1/2x-1/(2pi)intcos(u)du#

#=1/2x-1/(2pi)sin(pix)+C#