# Can't figure out this integral

MathWarrior
I am unsure how to solve this integral...

$\int5e^{\frac{t}{2}}sin(2t) dt$

Work I did:

$5\int e^{\frac{t}{2}}sin(2t) dt$

integration by parts wont work I don't think therefore I am stuck.

## Answers and Replies

nonaks

Try to let u = e^(t/2) and dv = sin2tdt, then the resulting integral will have a tail Integral (1/2cost(1/2e^(t/2)dt). Use the same process and eventually it will go back to the same integral on the problem. Then just use algebraic technique to solve the problem. Hope it helps. (im still at school at the moment, i'll figure this out when i reach home.)

MrWakaMaka

5∫et/2sin(2t)dt

Let:
u = sin(2t), therefore du = 2cos(2t)
vdv = et/2 and v = 2et/2

Hope this helps!

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MathWarrior

Was able to get to:

$5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}-4\int e^{\frac{t}{2}}cos(2t)$
$5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}-4[2cos(2t)e^{\frac{t}{2}} + 4\int e^{\frac{t}{2}}sin(2t) ]$
not sure what to do now.

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JHamm

I haven't checked your wwork but at this point you just need to rearrange the equation so your integral only appears on the left, you're so close
:-)

MathWarrior

$u = sin(2t)$
$du = 2cos(2t)$
$v = 2e^{\frac{t}{2}}$
$dv = e^{\frac{t}{2}}$

$5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}-4\int e^{\frac{t}{2}}cos(2t)$

$u = cos(2t)$
$du = -2sin(2t)$

$v=2e^{\frac{t}{2}}$
$dv=e^{\frac{t}{2}}$

$5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}-4[-2cos(2t)e^{\frac{t}{2}} + 4\int e^{\frac{t}{2}}sin(2t) ]$

$5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}+8cos(2t)e^{\frac{t}{2}} -16\int e^{\frac{t}{2}}sin(2t)$

What do I do with the:
$-16\int e^{\frac{t}{2}}sin(2t)$

JHamm

Move it to the left side, it might be easier to replace it with a 'j' or some such since all you have left to do is some algebra and you don't get distracted by the integral sign

MathWarrior

moving it over by adding to both sides I get :
$21 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}+8cos(2t)e^{\frac{t}{2}}$

this doesn't look right if i divide by 21 after this? I'm doing something wrong here I think.

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That's what you want to do, but you forgot the overall factor of 5 on the righthand side when you integrated by parts. So fix that first.

MathWarrior

That's what you want to do, but you forgot the overall factor of 5 on the righthand side when you integrated by parts. So fix that first.
I don't know how, or I would of already done it.

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You don't know how to multiply the righthand side by 5?

MathWarrior
You don't know how to multiply the righthand side by 5?

Why would you multiply by 5 instead of dividing by 5? To move stuff don't you do the inverse.

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I'm talking about the very beginning of your work.

What is ##2 e^{t/2}\sin 2t - 4 \int e^{t/2}\cos 2t\,dt##, which is what you got after the first integration by parts, equal to? Is it
$$\int e^{t/2}\sin 2t\, dt$$ or $$5 \int e^{t/2}\sin 2t\, dt?$$

MathWarrior
its equal to

$$5 \int e^{t/2}\sin 2t\, dt$$

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So tell me then, what is ##\int e^{t/2}\sin 2t\, dt## equal to after integrating by parts once?

MathWarrior
its equal to

$uv - \int v du$

where
$u = sin(2t)$
$du = 2cos(2t)$
$v = 2e^{\frac{t}{2}}$
$dv = e^{\frac{t}{2}}$

Giving

$2sin(2t)e^{\frac{t}{2}} - 4\int e^{\frac{t}{2}}cos(2t)$

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OK, in post 14, you said
$$2 e^{t/2}\sin 2t - 4 \int e^{t/2}\cos 2t\,dt = 5 \int e^{t/2}\sin 2t\, dt.$$
In post 14, you said
$$2 e^{t/2}\sin 2t - 4 \int e^{t/2}\cos 2t\,dt = \int e^{t/2}\sin 2t\, dt.$$
Since the lefthand sides are the same, you're saying
$$\int e^{t/2}\sin 2t\, dt = 5 \int e^{t/2}\sin 2t\, dt$$ Can you see the problem?

genericusrnme
just skimming over, I see it hasn't been suggested to use the exponential form of Sin(2t)

Any time I see an integral over a Cos or a Sin I find that it takes me a lot less time just to use the fact that
$Sin(at) = \frac{e^{i\ a\ t} - e^{-i\ a\ t}}{2\ i}$
Then you end up with two easy integrals of exopnents which you can then turn back into trig functions after everything has been done.

That might just be me though, most other people I know prefer u substitution.

lazyaditya
from applying by parts once you can see "cos x" and "sin x" thus use by parts once again on integral part !

MathWarrior
I can see that the two sides
$$\int e^{t/2}\sin 2t\, dt = 5 \int e^{t/2}\sin 2t\, dt$$
are not equal to each other.

So, since i found the original equation within the series of doing integration by parts over and over. And its not equal to $$5 \int e^{t/2}\sin 2t\, dt$$ I have to multiply the entire side by 5 ?

giving

$5 \int e^{\frac{t}{2}}sin(2t) = (5)2sin(2t)e^{\frac{t}{2}}+(5)8cos(2t)e^{\frac{t}{2}} -16(5)\int e^{\frac{t}{2}}sin(2t)$

which turns into

$5 \int e^{\frac{t}{2}}sin(2t) = 10sin(2t)e^{\frac{t}{2}}+40cos(2t)e^{\frac{t}{2}} -80\int e^{\frac{t}{2}}sin(2t)$
?

lazyaditya
just skimming over, I see it hasn't been suggested to use the exponential form of Sin(2t)

Any time I see an integral over a Cos or a Sin I find that it takes me a lot less time just to use the fact that
$Sin(at) = \frac{e^{i\ a\ t} - e^{-i\ a\ t}}{2\ i}$
Then you end up with two easy integrals of exopnents which you can then turn back into trig functions after everything has been done.

That might just be me though, most other people I know prefer u substitution.

ya this can be used or you can also do by considering sin(at)= Imaginary part of exponential [iat]

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giving

$5 \int e^{\frac{t}{2}}sin(2t) = (5)2sin(2t)e^{\frac{t}{2}}+(5)8cos(2t)e^{\frac{t}{2}} -16(5)\int e^{\frac{t}{2}}sin(2t)$

which turns into

$5 \int e^{\frac{t}{2}}sin(2t) = 10sin(2t)e^{\frac{t}{2}}+40cos(2t)e^{\frac{t}{2}} -80\int e^{\frac{t}{2}}sin(2t)$
?
Good! (But you have a sign error. The cosine term should be negative.)

Let's define ##I = 5\int e^{t/2}\sin 2t\,dt##, so you (should) have
$$I = 10e^{t/2}\sin 2t - 40 e^{t/2}\cos 2t - 16 I$$ Now just solve for I.

MathWarrior
I get

$\frac{10}{17} sin(2t) e^{\frac{t}{2}} - \frac{40}{17} e^{\frac{t}{2}}cos(2t)$

Thanks for the help, looks like this is equivalent.

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You got it! You can always check an answer by differentiating it and seeing if you recover the integrand.