How Do You Correctly Apply Integration by Parts to ∫-e^(2x)*sin(e^x) dx?

[shreddinglicks]

Homework Statement

I want to integrate

∫-e^(2x)*sin(e^x) dx

Homework Equations

∫uv'dx=uv - ∫u'v

The Attempt at a Solution

u = e^2x
du = 2*e^2x

dv = sin(e^x)
v = -cos(e^x)/e^x

e^(x)*cos(e^x) - 2∫e^(x)*cos(e^x) dx

e^(x)*cos(e^x) - 2*sin(e^x) + c

The solution I have doesn't have the two in the 2nd term of the answer. Not sure what I did wrong.

 

[Ray Vickson]

Homework Statement

I want to integrate

∫-e^(2x)*sin(e^x) dx

Homework Equations

∫uv'dx=uv - ∫u'v

The Attempt at a Solution

u = e^2x
du = 2*e^2x

dv = sin(e^x)
v = -cos(e^x)/e^x

e^(x)*cos(e^x) - 2∫e^(x)*cos(e^x) dx

e^(x)*cos(e^x) - 2*sin(e^x) + c

The solution I have doesn't have the two in the 2nd term of the answer. Not sure what I did wrong.

The integration of $v = \int \sin(e^x) \, dx$ is non-elementary, meaning that it cannot be expressed in terms of standard functions. In particular, if you compute $$\frac{d}{dx} \frac{-\cos(e^x)}{e^x}$$
you will see that you do not recover your hoped-for $dv = \sin(e^x) \, dx.$ So, you need another $dv$.

 

[Mark44]

You are omitting the dx factors in your work. They might not be some important in this problem, but doing this will definitely come back to bite you when you do integration by trig substitution.

In your work you have $u = e^{2x}$ and $dv = \sin(e^x) dx$ (dx added by me).

Can you think of another way to divide things between u and dv? Keep in mind that $e^{2x} = e^x \cdot e^x$.

 

[shreddinglicks]

You are omitting the dx factors in your work. They might not be some important in this problem, but doing this will definitely come back to bite you when you do integration by trig substitution.

In your work you have $u = e^{2x}$ and $dv = \sin(e^x) dx$ (dx added by me).

Can you think of another way to divide things between u and dv? Keep in mind that $e^{2x} = e^x \cdot e^x$.

I'm not sure if I follow what you're saying.

 

[shreddinglicks]

Are you saying I can factor out a e^x and leave it out as a constant?

 

[shreddinglicks]

The integration of $v = \int \sin(e^x) \, dx$ is non-elementary, meaning that it cannot be expressed in terms of standard functions. In particular, if you compute $$\frac{d}{dx} \frac{-\cos(e^x)}{e^x}$$
you will see that you do not recover your hoped-for $dv = \sin(e^x) \, dx.$ So, you need another $dv$.

I see what you mean.

 

[Mark44]

Are you saying I can factor out a e^x and leave it out as a constant?

No, that's not what I'm saying. The idea with integration by parts is figuring out how to divvy up the integrand into two pieces: u and dv.

Here are some possibilities (I am ignoring the minus sign in your problem for the time being):
$u = 1$, $dv = e^{2x}\sin(e^x)dx$ Obviously, that's a non-starter
$u = e^{2x}\sin(e^x)$, $dv = dx$ That's not a good choice, either
$u = e^{2x}$, $dv = \sin(e^x)dx$ This was your choice, which didn't work out

Considering my earlier hint, can you think of another way to divide up the integrand so as to choose dv so that it's actually something you can integrate? In integration by parts you want to choose dv so that it's the most complicated expression that you can still integrate. I can't really say much more without giving too much help.

 

[Dr Transport]

let y = e^x then the integral becomes
$$ -\int e^{2x}\sin(e^x) dx = -\int y\sin(y) dy $$ which is easily integrated by parts.

 

[shreddinglicks]

let y = e^x then the integral becomes
$$ -\int e^{2x}\sin(e^x) dx = -\int y\sin(y) dy $$ which is easily integrated by parts.

I like that, nice and simple.

 

[shreddinglicks]

No, that's not what I'm saying. The idea with integration by parts is figuring out how to divvy up the integrand into two pieces: u and dv.

Here are some possibilities (I am ignoring the minus sign in your problem for the time being):
$u = 1$, $dv = e^{2x}\sin(e^x)dx$ Obviously, that's a non-starter
$u = e^{2x}\sin(e^x)$, $dv = dx$ That's not a good choice, either
$u = e^{2x}$, $dv = \sin(e^x)dx$ This was your choice, which didn't work out

Considering my earlier hint, can you think of another way to divide up the integrand so as to choose dv so that it's actually something you can integrate? In integration by parts you want to choose dv so that it's the most complicated expression that you can still integrate. I can't really say much more without giving too much help.

dv = e^x * sin(e^x) dx
v = -cos(e^x)

u = e^x
du = e^x dx