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

Question

2632 views around the world

You can reuse this answer
Creative Commons License

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#