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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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