What is #int_0^(pi/4) cos^(2)8thetad theta#?

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Answer 1 · 2018-05-23T14:40:17

# pi/8#.

Recall that, #cos^2x=(1+cos2x)/2#.

#:. cos^2 8theta=1/2{1+cos(2xx8theta)}=1/2(1+cos16theta)#.

#:. int_0^(pi/4)cos^2 8theta d(theta)#,

#=1/2int_0^(pi/4)(1+cos16theta)d(theta)#,

#=1/2[theta+1/16sin16theta]_9^(pi/4)#,

#=1/2[{pi/4+1/16sin(16*pi/4)}-{0+1/16sin0}]#,

#=1/2{(pi/4+1/16sin4pi)-(0)}#,

#=1/2(pi/4+0)#,

#=pi/8#.