Integrate ∫e^7*sin(4x)

**UBCBOY** · February 18th 2013, 08:16 AM

Hi, I thought I had this question until I got to the very last part, I believe this question uses the repeated integration by parts method, is that right?

Evaluate the indefinite integral:

∫e^7x*sin(4x)

My work:
u = e^7x
u' = (1/7)e^7x
v = (-1/4)cos(4x)
v' = sin(4x)

∫e^7x*sin(4x) = (-1/4)e^7x*cos(4x) - ∫(-1/4)cos(4x)*(1/7)e^7x
.....................= (-1/4)e^7x*cos(4x) + (1/28) ∫cos(4x)*e^7x <-- integrate this indefinite integral using integration by parts?

So...
(1/28) ∫cos(4x)*e^7x
u = cos(4x)
u' = (-1/4)sin(4x)
v = (1/7)e^7x
v' = e^7x

And then I get...
(1/28) ∫cos(4x)*e^7x = (1/28)(cos(4x)*(1/7)e^7x + ∫(1/7)e^7x*(-1/4)sin(4x))
..............................= (1/196)(cos(4x) + (1/28^2)∫e^7x*sin(4x) <-- so this second term is now the same as the original equation, except for the constant of course, and I should just simplify, right? But I don't know the fractions make it look really complicated and I don't think I'm doing it right...

Does someone know where I went wrong and how to solve it properly?

**MINOANMAN** · February 18th 2013, 08:47 AM

UBCBOY

I FOUND e^(7x)[(7sin(4x)-4cos(4x))/65]+C
it is more simple ..check it
MINOAS

**veileen** · February 18th 2013, 08:59 AM

Uhm... seriously, use LaTeX. Sorry, I didn't read all you wrote. Anyway, $u'=7e^{7x}$ .

$I=\int e^{7x}\sin{4x}\, dx=\int \left (\frac{1}{7}e^{7x} \right )' \sin{4x}\, dx=$

$=\frac{1}{7}e^{7x}\sin{4x}-\int \frac{1}{7}e^{7x} (\sin{4x})'\, dx=\frac{1}{7}e^{7x}\sin{4x}-\int \frac{1}{7}e^{7x}\cdot 4\cos{4x}\, dx=$

$=\frac{1}{7}e^{7x}\sin{4x}-\frac{4}{7}\int e^{7x} \cos{4x}\, dx=\frac{1}{7}e^{7x}\sin{4x}-\frac{4}{7}\int \left (\frac{1}{7}e^{7x} \right )' \cos{4x}\, dx =$

$=\frac{1}{7}e^{7x}\sin{4x}-\frac{4}{7}\left ( \frac{1}{7}e^{7x}\cos{4x}- \int \frac{1}{7}e^{7x} (\cos{4x})'\, dx \right ) =$

$=\frac{1}{7}e^{7x}\sin{4x}-\frac{4}{49}e^{7x}\cos{4x}+\frac{4}{49}\int e^{7x}\cdot 4(-\sin{4x})\,dx=$

$=\frac{1}{7}e^{7x}\sin{4x}-\frac{4}{49}e^{7x}\cos{4x}-\frac{16}{49}\int e^{7x}\cdot \sin{4x}\,dx=$

$=\frac{1}{7}e^{7x}\sin{4x}-\frac{4}{49}e^{7x}\cos{4x}-\frac{16}{49}I\Rightarrow$

$\left ( 1+\frac{16}{49} \right )I=\frac{1}{7}e^{7x}\sin{4x}-\frac{4}{49}e^{7x}\cos{4x}\Rightarrow$

$65I=7e^{7x}\sin{4x}-4e^{7x}\cos{4x}\Rightarrow I=e^{7x}\left ( \frac{7}{65}\sin {4x}-\frac{4}{65} \cos{4x} \right )+C$

**UBCBOY** · February 18th 2013, 09:10 AM

Thank you guys. Is LaTex a program for solving calculus?

**ILikeSerena** · February 18th 2013, 10:13 AM

Originally Posted by UBCBOY

Thank you guys. Is LaTex a program for solving calculus?

Hey UBCBOY!

LaTeX is a typesetting format.
If you click Reply With Quote on veileen's post, you'll see how it works.
Basically you type your formulas like you'd usually would, then add [ TEX ] ... [ /TEX ] markers around it, and presto!
A nicely formatted formula.
If you click Go Advanced, you'll see a quick button with a $\sum$ on it that facilitates this.

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