Int_theta^pi/2 1/3+Sin^2theta?

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Answer 1 · 2018-05-15T00:05:40

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#int_(theta)^(pi/2) 1/3+Sin^2theta*d(theta)#

#int_(theta)^(pi/2) 1/3+[1/2-cos[2(theta)]/(2)]*d(theta)#

#=[1/3theta+1/2theta-1/4sin(2theta)]_theta^(pi/2)#

#=[1/3(pi/2)+1/2(pi/2)-1/4sin(2(pi/2))]-[1/3theta+1/2theta-1/4sin(2theta)]#

#=[(pi/6)+(pi/4)-1/4sin(pi)]-[1/3theta+1/2theta-1/4sin(2theta)]#

#=[(pi/6)+(pi/4)]-[1/3theta+1/2theta-1/4sin(2theta)]#