# Find int_0^(π/2) e^x sin xdx ?

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Jim S Share
Mar 9, 2018

int_0^(π/2)e^xsinxdx=(e^(π/2)+1)/2

#### Explanation:

I=int_0^(π/2)e^xsinxdx

We will use integration by parts

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

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