How do you find the antiderivative of #sin(pix) dx#?

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How do you find the antiderivative of #sin(pix) dx#?

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Answer 1 · 2017-05-01T20:35:50

#" "intsin(pix)dx=-1/picos(pix)+C#

Explanation:

Let
#" "#
#u=cos(pix)#
#" "#
#du=-pisin(pix)dx#
#" "#
#-(du)/pi=sin(pix)dx#
#" "#
#intsin(pix)dx=int-(du)/pi=-1/piintdu=-1/pi(u)+C#
#" "#
Therefore,#" "intsin(pix)dx=-1/picos(pix)+C#

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How do you find the antiderivative of #sin(pix) dx#?

1 Answer

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