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#