Question #a888b

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Answer 1 · 2017-10-22T06:53:59

The answer is $= 2$ $=2$

We need

$\int cos x d x = sin x + C$ $intcosxdx=sinx+C$

$\int sin 2 x d x = - \frac{1}{2} cos 2 x + C$ $intsin2xdx=-1/2cos2x+C$

Therefore,

$\int_{0}^{\frac{π}{2}} (cos x + sin 2 x) = {[sin x - \frac{1}{2} cos 2 x]}_{0}^{\frac{π}{2}}$ $int_0 ^(pi/2)(cosx+sin2x)=[sinx-1/2cos2x]_0^(pi/2)$

$= (sin (\frac{π}{2}) - \frac{1}{2} cos π) - (sin 0 - \frac{1}{2} cos 0)$ $=(sin(pi/2)-1/2cospi)-(sin0-1/2cos0)$

$= 1 + \frac{1}{2} - 0 + \frac{1}{2}$ $=1+1/2-0+1/2$

$= 2$ $=2$