- Forums
- Homework Help
- Calculus and Beyond Homework Help
- Thread startershreddinglicks
- Start date
- Tags
- IntegrationIntegration by partsparts
In summary, the student attempted to integrate a function but ran into trouble because the function was non-elementary. They attempted to find another solution but failed.
- #1
shreddinglicks
- 212
- 6
Homework Statement
I want to integrate
∫-e^(2x)*sin(e^x) dx
Homework Equations
∫uv'dx=uv - ∫u'v
The Attempt at a Solution
u = e^2x
du = 2*e^2x
dv = sin(e^x)
v = -cos(e^x)/e^x
e^(x)*cos(e^x) - 2∫e^(x)*cos(e^x) dx
e^(x)*cos(e^x) - 2*sin(e^x) + c
The solution I have doesn't have the two in the 2nd term of the answer. Not sure what I did wrong.
Physics news on Phys.org
- #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
- 10,704
- 1,722
shreddinglicks said:
Homework Statement
I want to integrate
∫-e^(2x)*sin(e^x) dx
Homework Equations
∫uv'dx=uv - ∫u'v
The Attempt at a Solution
u = e^2x
du = 2*e^2xdv = sin(e^x)
v = -cos(e^x)/e^xe^(x)*cos(e^x) - 2∫e^(x)*cos(e^x) dx
e^(x)*cos(e^x) - 2*sin(e^x) + c
The solution I have doesn't have the two in the 2nd term of the answer. Not sure what I did wrong.
The integration of ##v = \int \sin(e^x) \, dx## is non-elementary, meaning that it cannot be expressed in terms of standard functions. In particular, if you compute $$\frac{d}{dx} \frac{-\cos(e^x)}{e^x}$$
you will see that you do not recover your hoped-for ##dv = \sin(e^x) \, dx.## So, you need another ##dv##.
- #3
Mark44
- 37,617
- 9,846
You are omitting the dx factors in your work. They might not be some important in this problem, but doing this will definitely come back to bite you when you do integration by trig substitution.
In your work you have ##u = e^{2x}## and ##dv = \sin(e^x) dx## (dx added by me).
Can you think of another way to divide things between u and dv? Keep in mind that ##e^{2x} = e^x \cdot e^x##.
- #4
shreddinglicks
- 212
- 6
Mark44 said:
You are omitting the dx factors in your work. They might not be some important in this problem, but doing this will definitely come back to bite you when you do integration by trig substitution.
In your work you have ##u = e^{2x}## and ##dv = \sin(e^x) dx## (dx added by me).
Can you think of another way to divide things between u and dv? Keep in mind that ##e^{2x} = e^x \cdot e^x##.
I'm not sure if I follow what you're saying.
- #5
shreddinglicks
- 212
- 6
Are you saying I can factor out a e^x and leave it out as a constant?
- #6
shreddinglicks
- 212
- 6
Ray Vickson said:
The integration of ##v = \int \sin(e^x) \, dx## is non-elementary, meaning that it cannot be expressed in terms of standard functions. In particular, if you compute $$\frac{d}{dx} \frac{-\cos(e^x)}{e^x}$$
you will see that you do not recover your hoped-for ##dv = \sin(e^x) \, dx.## So, you need another ##dv##.
I see what you mean.
- #7
Mark44
Mentor
Insights Author
- 37,617
- 9,846
shreddinglicks said:
Are you saying I can factor out a e^x and leave it out as a constant?
No, that's not what I'm saying. The idea with integration by parts is figuring out how to divvy up the integrand into two pieces: u and dv.
Here are some possibilities (I am ignoring the minus sign in your problem for the time being):
##u = 1##, ##dv = e^{2x}\sin(e^x)dx## Obviously, that's a non-starter
##u = e^{2x}\sin(e^x)##, ##dv = dx## That's not a good choice, either
##u = e^{2x}##, ##dv = \sin(e^x)dx## This was your choice, which didn't work out
, can you think of another way to divide up the integrand so as to choose dv so that it's actually something you can integrate? In integration by parts you want to choose dv so that it's the most complicated expression that you can still integrate. I can't really say much more without giving too much help.
- #8
Dr Transport
Science Advisor
Insights Author
Gold Member
- 2,606
- 800
let [itex] y = e^x[/itex] then the integral becomes
$$ -\int e^{2x}\sin(e^x) dx = -\int y\sin(y) dy $$ which is easily integrated by parts.
- #9
shreddinglicks
- 212
- 6
Dr Transport said:
let [itex] y = e^x[/itex] then the integral becomes
$$ -\int e^{2x}\sin(e^x) dx = -\int y\sin(y) dy $$ which is easily integrated by parts.
I like that, nice and simple.
- #10
shreddinglicks
- 212
- 6
Mark44 said:
No, that's not what I'm saying. The idea with integration by parts is figuring out how to divvy up the integrand into two pieces: u and dv.
Here are some possibilities (I am ignoring the minus sign in your problem for the time being):
Considering my earlier hint
##u = 1##, ##dv = e^{2x}\sin(e^x)dx## Obviously, that's a non-starter
##u = e^{2x}\sin(e^x)##, ##dv = dx## That's not a good choice, either
##u = e^{2x}##, ##dv = \sin(e^x)dx## This was your choice, which didn't work out, can you think of another way to divide up the integrand so as to choose dv so that it's actually something you can integrate? In integration by parts you want to choose dv so that it's the most complicated expression that you can still integrate. I can't really say much more without giving too much help.
dv = e^x * sin(e^x) dx
v = -cos(e^x)
u = e^x
du = e^x dx
1. What is integration by parts?
Integration by parts is a method of integration used to solve integrals that are in the form of a product of two functions. It is based on the product rule of differentiation, where the integral of the product f(x)g(x) can be found by integrating f(x) and differentiating g(x).
2. When should I use integration by parts?
Integration by parts is typically used when the integral cannot be solved by other methods such as substitution or partial fractions. It is also useful when the integrand contains a function that is easy to differentiate, but difficult to integrate.
3. How do I choose which function to integrate and which to differentiate?
The choice of which function to integrate and which to differentiate is based on the acronym "LIATE", which stands for "Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential" functions. Generally, the function that comes first in this order is chosen to be integrated, while the second function is differentiated.
4. Can integration by parts be used for definite integrals?
Yes, integration by parts can be used for definite integrals. In this case, the limits of integration are applied to the final integral after integration by parts has been performed.
5. Are there any common mistakes to avoid when using integration by parts?
One common mistake to avoid when using integration by parts is forgetting the "+ C" at the end of the integration. It is also important to pay attention to the choice of u and dv, making sure they are easy to integrate and differentiate respectively. Additionally, it is important to check for algebraic errors when simplifying the final integral.
Similar threads
Differentiate the given integral
- Calculus and Beyond Homework Help
- Replies
- 15
- Views
- 783
Arc length of vector function - the integral seems impossible
- Calculus and Beyond Homework Help
- Replies
- 2
- Views
- 837
Solve ##\int\frac{e^{-x}}{x^2}dx## and ##\int \frac{e^{-x}}{x}dx##
- Calculus and Beyond Homework Help
- Replies
- 7
- Views
- 704
Solve the given problem that involves integration
- Calculus and Beyond Homework Help
- Replies
- 4
- Views
- 738
Determine whether ## S[y] ## has a maximum or a minimum
- Calculus and Beyond Homework Help
- Replies
- 18
- Views
- 1K
Find ##f(x)## in the problem involving integration
- Calculus and Beyond Homework Help
- Replies
- 3
- Views
- 791
Evaluate ##\int_{-\infty}^{\infty} e^{-|x|}\delta(x^2 +2x -3) dx##
- Calculus and Beyond Homework Help
- Replies
- 7
- Views
- 1K
First Order Diffy Q Problem with Bernoulli/Integrating Factors
- Calculus and Beyond Homework Help
- Replies
- 4
- Views
- 918
Solve the given first order differential equation
- Calculus and Beyond Homework Help
- Replies
- 8
- Views
- 759
Integration problem using inscribed rectangles
- Calculus and Beyond Homework Help
- Replies
- 14
- Views
- 236
- Forums
- Homework Help
- Calculus and Beyond Homework Help