Integration by parts with e and sine

[Yae Miteo]

Homework Statement

Evaluate the integral.

Homework Equations

$$\int e^{2x} sin(3x) dx$$

The Attempt at a Solution

I began by using integration by parts.

$$u = sin(3x)$$

$$v = \frac {e^{2x}} {2}$$

$$du = 3 cos(3x)$$

$$dv = e^{2x} dx$$

but I get stuck after that because the integral keeps breaking down into more and more integrals, and I never arrive at the correct answer. I also may be making this too complicated. Any ideas?

 

[STEMucator]

Homework Statement

Evaluate the integral.

Homework Equations

[tex]\int e^2{x} sin(3x) dx [/int]<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> I began by using integration by parts.<br /> <br /> $$u = sin(3x) v = \frac {e^2x} {2} du = 3 cos(3x) dv = e^2x dx$$<br /> <br /> but I get stuck after that because the integral keeps breaking down into more and more integrals, and I never arrive at the correct answer. I also may be making this too complicated. Any ideas?[/tex]

[tex] <br /> I believe you meant to write your integral as:<br /> <br /> $$I = \int e^{2x} sin(3x) dx $$<br /> <br /> Notice I've called the above integral $I$, the reason for this will become apparent soon.<br /> <br /> Make the substitution $u = sin(3x)$ and $dv = e^{2x} dx$. What is the result of the following integral:<br /> <br /> $$uv - \int v du$$[/tex]

 

[Mark44]

Homework Statement

Evaluate the integral.

Homework Equations

$$\int e^{2x} sin(3x) dx$$

The Attempt at a Solution

I began by using integration by parts.

$$u = sin(3x)$$

$$v = \frac {e^{2x}} {2}$$

$$du = 3 cos(3x)$$

$$dv = e^{2x} dx$$

but I get stuck after that because the integral keeps breaking down into more and more integrals, and I never arrive at the correct answer. I also may be making this too complicated. Any ideas?

You need to do integration by parts once more. If you use similar substitutions, you'll end up with another integral whose integrand is e^2xsin(3x). At that point, you can solve algebraically for this integral.

 

[Feodalherren]

When you do parts twice, the same function will pop up inside the integral on both sides:

$\int e^{2x}Sin3x dx = \frac{1}{2} e^{2x} Sin3x - \frac{3}{4}e^{2x} Cos3x - \frac{9}{4} \int e^{2x} Sin3x dx$

Notice how the integrand is the same on both sides, this means that you can move your 9/4 over to the left using algebra. If you call your integral I in this case, all you need to do is solve for I.

You should put this special case in your memory bank - it always happens with trig functions and e since neither will be reduced by using parts.

 

[Saitama]

I like to do these problems in the following way as it gives the answer quickly.

Consider:
$$A=\int e^{2x}\cos (3x)\,dx$$
$$B=\int e^{2x}\sin (3x)\,dx$$
Hence,
$$A+iB=\int e^{2x}e^{i3x}\,dx=\int e^{(2+3i)x}\,dx=\frac{e^{(2+3i)x}}{2+3i}+C$$
To obtain the given integral, compare the imaginary parts.