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How do I evaluate #int_0^pisin(x)dx?#

1 Answer

Apr 22, 2018

#2#

Explanation:

Given: #int_0^pisin(x) \ dx#.

#=[-cos(x)]""_0^pi#

#=-cos(pi)-(-cos(0))#

#=-cos(pi)+cos(0)#

#=-(-1)+1#

#=1+1#

#=2#

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