How do you evaluate the definite integral #int sin3x# from #[0,pi]#?

2 Answers
sjc
Aug 29, 2017

#2/3#

Explanation:

#int_0^(pi)sin3xdx#

now#d/(dx)(cos3x)=-3sin3x#

#:.int_0^(pi)sin3xdx=[-1/3cos3x]_0^pi#

#=-1/3{[cos3x]^pi-[cos3x]_0}#

#=-1/3(cos3pi-cos0)#

#=-1/3(-1-1)#

#-1/3xx-2#

#=2/3#

Steve M
Aug 29, 2017

# int_0^pi sin3x = 2/3#

Explanation:

Using:

# d/dx{cosax}=-asinax#

Then:

# int_0^pi sin3x = [-1/3cos3x]_0^pi #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -1/3(cos3pi - cos 0)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -1/3(-1- 1)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2/3#

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