Evaluate $int_0^pi(2+ sinx)dx$ ?

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Answer 1 · 2016-11-29T18:18:41

$int_0^pi(2+ sinx)dx = 2pi+2$

You should learn that;

$d/dx sinx = cosx iff int cosx dx = sinx + c$
$d/dx cosx = -sinx iff int sinx = -cosx + c$

So

$int_0^pi(2+ sinx)dx = [2x - cosx]_0^pi$
$\ = (2pi - cospi) - (2*0-cos0)$
$\ = (2pi - (-1)) - (0-1)$
$\ = 2pi+1 +1$
$\ = 2pi+2$

Evaluate int_0^pi(2+ sinx)dx int_0^pi(2+ sinx)dx?