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Question #a888b

1 Answer

Oct 22, 2017

The answer is #=2#

Explanation:

We need

#intcosxdx=sinx+C#

#intsin2xdx=-1/2cos2x+C#

Therefore,

#int_0 ^(pi/2)(cosx+sin2x)=[sinx-1/2cos2x]_0^(pi/2)#

#=(sin(pi/2)-1/2cospi)-(sin0-1/2cos0)#

#=1+1/2-0+1/2#

#=2#

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