Find $int_0^(π/2) e^x sin xdx$ ?

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Answer 1 · 2018-03-09T01:03:46

$int_0^(π/2)e^xsinxdx=(e^(π/2)+1)/2$

$I=int_0^(π/2)e^xsinxdx$

$int_0^(π/2)e^xsinxdx=int_0^(π/2)(e^x)'sinxdx$ $=$

$[e^xsinx]_0^(π/2) - int_0^(π/2)e^x(sinx)'dx$

$[e^xsinx]_0^(π/2) - int_0^(π/2)e^xcosxdx$

$[e^xsinx]_0^(π/2) - int_0^(π/2)(e^x)'cosxdx$

$[e^xsinx]_0^(π/2) - [e^xcosx]_0^(π/2)+int_0^(π/2)e^x(-sinx)dx$

$[e^xsinx]_0^(π/2) - [e^xcosx]_0^(π/2)-int_0^(π/2)e^xsinxdx$

If $int_0^(π/2)e^xsinxdx=J$

then

$I+J=2int_0^(π/2)e^xsinxdx$

$2int_0^(π/2)e^xsinxdx=[e^xsinx]_0^(π/2) - [e^xcosx]_0^(π/2)$

$int_0^(π/2)e^xsinxdx=([e^xsinx]_0^(π/2) - [e^xcosx]_0^(π/2))/2$

$int_0^(π/2)e^xsinxdx=(e^(π/2)sin(π/2)-e^0sin0 - e^(π/2)cos(π/2)+e^0cos0)/2$

$int_0^(π/2)e^xsinxdx=(e^(π/2)+1)/2$

Find int_0^(π/2) e^x sin xdxint_0^(π/2) e^x sin xdx ?