What is the integral of #int sin (pi x) dx#?

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Answer 1 · 2016-02-11T01:04:57

#(-cos(pix))/pi+C#

Noting that #intsin(x)dx=-cos(x)+C#,

Let #u=pix#, so #du=pidx#.

From here, we have

#=intsin(u)(1/pi)pidx#

Bring the constant out of the expression and replace #pidx# with #du#:

#=1/piintsin(u)du=-(1/pi)cos(u)+C#

Substituting #u=pix# back in, we get a final answer of

#(-cos(pix))/pi+C#