Evaluate the following Integral (II)

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Now, put $x=0$ to get $C$. So, $$B=\frac{e^x(\sin(x)-\cos(x))}{2}-\frac{1}{2}$$So,$$\int e^x\sin(x)\,dx=\frac{e^x(\sin(x)-\cos(x))}{2}-\frac{1}{2}$$
  • #1
shamieh
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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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  • #2
shamieh said:
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.
 
  • #3
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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  • #4
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
 
  • #5
so you would get \(\displaystyle I = \frac{e^x sinx - e^x cosx}{2} + C \)correct?
 
  • #6
shamieh said:
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.
 
  • #7
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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1. What is the purpose of evaluating an integral?

Evaluating an integral is used to determine the area under a curve or the accumulated value of a function over a given interval. It is also used in physics and engineering to calculate quantities such as displacement, velocity, and acceleration.

2. How do you evaluate an integral?

To evaluate an integral, you can use techniques such as substitution, integration by parts, or partial fractions. You can also use tables of integrals or computer software to help solve more complicated integrals.

3. What are the limits of integration?

Limits of integration specify the interval over which you want to evaluate the integral. They are typically represented by numbers or variables and are written as the lower and upper bounds of the integral. For example, in the integral ∫05 x dx, the limits of integration are 0 (lower bound) and 5 (upper bound).

4. Can an integral have multiple solutions?

Yes, an integral can have multiple solutions. This is because integrals are defined as an infinite sum of infinitely many infinitesimal rectangles, and there can be different ways to sum up these rectangles to get the same final area or value.

5. What is the relationship between integrals and derivatives?

The relationship between integrals and derivatives is known as the fundamental theorem of calculus. It states that the integral of a function is the inverse operation of its derivative. In other words, if you integrate a function and then take the derivative of the resulting function, you will get back the original function.

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