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

by using par partes (obviously). Thanks

2 Answers
Feb 19, 2018

=(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

Feb 20, 2018

{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