How do you evaluate the definite integral #int cos2theta# from #[0,pi/4]#?

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Answer 1 · 2017-02-18T13:33:19

#1/2#

#int_0^(pi/4)cos2thetad theta#

now

#d/(d theta)(sin2 theta)=2cos2 theta#

so

#int_0^(pi/4)cos2thetad theta=1/2[sin2 theta]_0^(pi/4)#

#1/2[sin2 theta]_0^(pi/4)=1/2[sin 2(pi/4)-cancel(sin (2xx0))]#

#=1/2sin(pi/2)=1/2xx1=1/2#