How do you evaluate the definite integral #int sin(4x)# from #[0,pi/8]#?

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Answer 1 · 2018-04-04T03:04:44

#1/4#

This is a fairly common integral.

#intsin(ax)dx=-1/acos(ax)# where #a# is any constant.

So,

#int_0^(pi/8)sin(4x)dx=-1/4cos4x|_0^(pi/8)#

Evaluating, we get

#-1/4cos((4pi)/8)-(-1/4cos(0))=-1/4cos(pi/2)+1/4cos(0)=1/4#

As #cos0=1#