Question

Would you help me? `int_0^(pi/2)(e^(2x)*sinx)dx`

by using par partes (obviously). Thanks

Answer 1

`=(2e^(pi)+1)/5`

Explanation:

this requires integration by parts as follows. The limits will be omitted until the very end

`int(e^(2x)sinx)dx`

`color(red)(I=intu(dv)/(dx)dx)=uv-intv(du)/(dv)dx`

`u=e^(2x)=>du=2e^(2x)dx`

`(dv)/(dx)=sinx=>v=-cosx`

`color(red)(I)=-e^(2x)cosx+int2e^(2x)cosxdx`

the second integral is also done by parts

`u=2e^(2x)=>du=4e^(2x)dx`

`(dv)/(dx)=cosx=>v=sinx`

`color(red)(I)=-e^(2x)cosx+[2e^(2x)sinx-int4e^(2x)sinxdx]`

`color(red)(I)=-e^(2x)cosx+2e^(2x)sinx-4color(red)(I)`

`:.5I=e^(2x)(2sinx-cosx)`

`I=(e^(2x)(2sinx-cosx))/5`

now put the limits in

`I=[(e^(2x)(2sinx-cosx))/5]_0^(pi/2)`

`=(e^pi((2sin(pi/2)-cos(pi/2)))/5)-(e^(0)(sin0-cos0)/5)`

`1/5e^pi[2-0]+1/5[ -0+1]`

`=(2e^(pi)+1)/5`

Answer 2

`{2e^pi+1}/5`

Explanation:

While the answer already provided is perfect, I just wanted to point out an easier way to arrive at the same answer using a slightly more advanced approach - that via complex numbers.

We start with the famous relation

`e^{ix} = cos(x)+i sin(x)`

where `i=sqrt{-1}`, and note that this means that

`sin(x) = \Im(e^{ix}) implies e^{2x}sin(x) = Im(e^{(2+i}x))`

where `Im` denotes the imaginary part.

So

`int_0^{pi/2} e^{2x}sin(x) dx = Im(int_0^{pi/2}e^{(2+i)x} dx)`
`=Im(e^{(2+i)x}/{2+i}|_0^{pi/2}) = Im({e^pi e^{ipi/2}-1}/{2+i})`
`=Im({ie^pi -1}/{2+i} times {2-i}/{2-i}) = 1/5 Im((-1+ie^pi)(2-i))`
`= 1/5((-1)times (-1)+e^pi times 2) = {2e^pi+1}/5`