Evaluate the following Integral (II)

[shamieh]

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

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

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

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

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

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

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

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

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

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

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

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

 

[Prove It]

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

 

[shamieh]

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?

 

[MarkFL]

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]

so you would get \(\displaystyle I = \frac{e^x sinx - e^x cosx}{2} + C \)correct?

 

[MarkFL]

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.

 

[Saitama]

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:
$$B=\frac{e^x(\sin(x)-\cos(x))}{2}+C$$