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Evaluate the following Integral (II)

shamieh

Active member
Sep 13, 2013
539
Evaluate the following integral

I think this is a recursion type problem, but I'm not quite sure. Again, I could be going about this problem horribly. Just need someone to check to see if this is the correct answer or if I'm even close to doing this problem right.


\(\displaystyle \int sin(x) e^x \, dx\)

\(\displaystyle u = sinx\)
\(\displaystyle du = cosx dx\)

\(\displaystyle dv = e^x\)
\(\displaystyle v = e^x\)

\(\displaystyle sinx e^x - \int e^x cosx \,dx\)

\(\displaystyle u = e^x \)
\(\displaystyle du = e^x \)

\(\displaystyle dv = -sinx\)
\(\displaystyle v = cosx\)

\(\displaystyle sinx e^x - e^x cos x - \int cosx e^x dx\)

\(\displaystyle u = cosx\)
\(\displaystyle du = -sinx\)

\(\displaystyle dv = e^x\)
\(\displaystyle v = e^x\)

\(\displaystyle I = sinx e^x - cosx e^x - cosx e ^x + \int e^x sinx \, dx\)

\(\displaystyle 2I = sinx e^x - 2cosx e^x\)

\(\displaystyle I = \frac{sinx e^x - 2cosx e^x}{2}\)
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Evaluate the following integral

I think this is a recursion type problem, but I'm not quite sure. Again, I could be going about this problem horribly. Just need someone to check to see if this is the correct answer or if I'm even close to doing this problem right.


\(\displaystyle \int sin(x) e^x \, dx\)

\(\displaystyle u = sinx\)
\(\displaystyle du = cosx dx\)

\(\displaystyle vu = e^x\)
\(\displaystyle v = e^x\)

\(\displaystyle sinx e^x - \int e^x cosx \,dx\)

\(\displaystyle u = e^x \)
\(\displaystyle du = e^x \)

\(\displaystyle vu = -sinx\)
\(\displaystyle v = cosx\)

\(\displaystyle sinx e^x - e^x cos x - \int cosx e^x dx\)

\(\displaystyle u = cosx\)
\(\displaystyle du = -sinx\)

\(\displaystyle vu = e^x\)
\(\displaystyle v = e^x\)

\(\displaystyle I = sinx e^x - cosx e^x - cosx e ^x + \int e^x sinx \, dx\)

\(\displaystyle 2I = sinx e^x - 2cosx e^x\)

\(\displaystyle I = \frac{sinx e^x - 2cosx e^x}{2}\)
Start by writing [tex]\displaystyle \begin{align*} I = \int{e^x\sin{(x)}\,dx} \end{align*}[/tex] and apply Integration by Parts twice, keeping the [tex]\displaystyle \begin{align*} e^x \end{align*}[/tex] term either as "u" or as "dv" in BOTH. You will find that [tex]\displaystyle \begin{align*} I \end{align*}[/tex] can be written in terms of [tex]\displaystyle \begin{align*} I \end{align*}[/tex], and thus can be solved.
 

shamieh

Active member
Sep 13, 2013
539
so you're saying

\(\displaystyle \int sin(x) e^x
\)
\(\displaystyle u = sinx\)
\(\displaystyle du = cosx\)

\(\displaystyle dv = e^x \)
\(\displaystyle v = e^x \)

\(\displaystyle sinx e^x - \int e^x cos x\)

\(\displaystyle u = cos x\)
\(\displaystyle du = - sinx\)

\(\displaystyle dv = e^x\)
\(\displaystyle v = e^x \)

\(\displaystyle sinx e ^x - cosx e^x + \int e^x sinx\)

\(\displaystyle I = e^x sinx - cosx e^x + \int e^x sinx\)

so now do I just subtract the first \(\displaystyle e^x sinx \) and the \(\displaystyle e^x sinx \)to the otehr side?
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You've made a mistake with a sign, and have been a bit lax with notation concerning differentials. I would take Prove It's advice and begin with:

\(\displaystyle I=\int e^x\sin(x)\,dx\)

Use integration by parts where:

\(\displaystyle u=\sin(x)\,\therefore\,du=\cos(x)\,dx\)

\(\displaystyle dv=e^x\,\therefore\,v=e^x\)

And we have:

\(\displaystyle I=e^x\sin(x)-\int e^x\cos(x)\,dx\)

Use integration by parts again where:

\(\displaystyle u=\cos(x)\,\therefore\,du=-\sin(x)\,dx\)

\(\displaystyle dv=e^x\,\therefore\,v=e^x\)

And we have:

\(\displaystyle I=e^x\sin(x)-\left(e^x\cos(x)+\int e^x\sin(x)\,dx \right)\)

Since \(\displaystyle I=\int e^x\sin(x)\,dx\), and distributing the negative sign, we get:

\(\displaystyle I=e^x\sin(x)-e^x\cos(x)-I\)

Now solve for $I$ and append the constant of integration. :D
 

shamieh

Active member
Sep 13, 2013
539
so you would get \(\displaystyle I = \frac{e^x sinx - e^x cosx}{2} + C \)correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
so you would get \(\displaystyle I = \frac{e^x sinx - e^x cosx}{2} + C \)correct?
Yes, that's correct. :D

When you begin your study of ordinary differential equations, you will find a technique that you can use to find anti-derivatives of this type without using integration by parts, which relies on the methods used to solve inhomogeneous linear ODEs.
 

Pranav

Well-known member
Nov 4, 2013
428
Alternative approach:

Let
$$B=\int e^x\sin(x) \,dx$$
$$A=\int e^x\cos(x)\, dx$$
We get,
$$A+iB=\int e^{x(1+i)}\,dx =\frac{e^{x(1+i)}}{1+i}+C=\frac{e^x(\cos(x)+i\sin(x))(1-i)}{2}+C$$

Since we need the imaginary part, hence:
$$B=\frac{e^x(\sin(x)-\cos(x))}{2}+C$$
 
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