How to find the integral of πcosπx?

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Answer 1 · 2018-05-10T13:47:00

#sin(pix)+C#

Given: #intpicos(pix) \ dx#.

Take out the constant.

#=piintcos(pix) \ dx#

Let's calculate #intcos(pix) \ dx#.

Let #u=pix,:.du=pi \ dx,dx=(du)/pi#.

#=intcosu \ (du)/pi#

Take out the constant again.

#=1/piintcosu \ du#

#=1/pisinu+C#

Putting that back, we get:

#=color(red)cancelcolor(black)(pi)*1/(color(red)cancelcolor(black)(pi))sinu+C#

#=sinu+C#

Putting back #u=pix#, we get the final answer:

#=sin(pix)+C#