How do you evaluate the definite integral $int sinxdx$ from $[0, pi/2]$ ?

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Answer 1 · 2017-02-14T09:00:42

$1$

$int_0^(pi/2)sinxdx$

now $d/(dx)(cosx)=-sinx$

so $intsinxdx=-cosxdx$

$:.int_0^(pi/2)sinxdx=[-cos]_0^(pi/2)$

$=-[cos]_0^(pi/2)=-{cancel((cos(pi/2)))^(=0)-cos0)}$

$=- - 1=1$

How do you evaluate the definite integral int sinxdxint sinxdx from [0, pi/2][0, pi/2]?