Question:
Integrate e^2x(sin x) ?
ndavos
2007-12-21 08:25:49 UTC
Use integration by parts to find the integral of e^2x(sin x)

Can someone help me as I seem to just go round in circles? Thanks.
Five answers:
mohanrao d
2007-12-21 08:41:18 UTC
∫e^(2x)(sinx)dx



integrate by parts



let u = e^2x : du = 2e^2x dx



let dv = sinx : v = -cosx



∫e^(2x) sinx dx = uv - ∫v du



∫e^(2x)sinx dx = -e^(2x)cosx + 2∫e^(2x)cosx dx



again integrate by parts



let u1 = e^(2x) : du1 = 2e^(2x) dx



let dv1 = cosx : v = sinx



∫e^(2x)sinx dx = -e^(2x)cosx + 2[e^(2x)sinx - 2∫e^(2x)sinx dx



∫e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx - 4∫e^(2x)sinx dx



add 4∫e^(2x)sinx dx on both sides to eliminate integral on RHS



5∫e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx



divide by 5 throughout



∫e^(2x)sinx dx = (-1/5)e^(2x)cosx + (2/5)e^(2x)sinx + c
james w
2007-12-21 08:35:12 UTC
I did integration by parts twice.



I got 1/5 ((2e^2x) sin x -(e^2x) cos x dx)



The second time you integrate, you will get the same thing again, but it will have a minus sign. Add it to the original and divide by the scalar ( which is 5).
Madhukar
2007-12-21 09:13:19 UTC
Use the formula

∫ e^ax * sin (bx) dx = [e^ax / √(a^2 + b^2)] * [a sin (bx) - b cos (bx)] + c



In the given problem, a = 2 and b = 1.

=> ∫ e^2x * sin (x) dx

= [e^2x / √(2^2 + 1^2)] * [2 sin (x) - cos (x)] + c

= [e^2x / 5] * [2 sin (x) - cos (x)] + c
Alexander
2007-12-21 08:43:23 UTC
You are supposed to go in circles. Until you figure out the way out, that is. Which is kind of tricky:



Let denote f(x) = ∫exp(2x) sin(x) dx



f(x) = ∫exp(2x) sin(x) dx =

= - ∫exp(2x) d[cos(x)] =

by parts

= -[ exp(2x) cos(x) - ∫cos(x) d[exp(2x)] ]=

= -[ exp(2x) cos(x) - 2∫exp(2x) cos(x) dx] ]=

= -[ exp(2x) cos(x) - 2∫exp(2x) d[sin(x)] ]=

by parts agian

= -[exp(2x) cos(x) - 2[exp(2x) sin(x) - ∫sin(x) d[exp(2x)] ]] =

= -[exp(2x)[cos(x) - 2sin(x)] + 4 ∫sin(x) exp(2x) dx ] =

now the tricky part:

= -[exp(2x)[cos(x) - 2sin(x)] + 4f(x)]



or

f(x) = -[exp(2x)[cos(x) - 2sin(x)] + 4f(x)]

5f(x) = -exp(2x)[cos(x) - 2sin(x)]

f(x) = -1/5 exp(2x)[cos(x) - 2sin(x)]
Nur S
2007-12-21 08:34:39 UTC
remember that , sin(x) = (e^ix - e^-ix)/2i

it will make things easier


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
Loading...