Definite integral #int_0^(pi/8)cos^2(2x)dx# ?

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#int_0^(pi/8)cos^2(2x)dx#

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Answer 1 · 2018-02-07T17:44:37

# 1/16(pi+2)#.

We have, #1+cos2theta=2cos^2theta#.

#:. int_0^(pi/8)cos^2 2xdx#,

#=int_0^(pi/8)(1+cos4x)/2dx#,

#=1/2[x+1/4sin4x]_0^(pi/8)#,

#=1/2[pi/8+1/4sin(4*pi/8)-0]#,

#=1/2(pi/8+1/4*1)#,

#=1/16(pi+2)#.