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^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.
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##.
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##.
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.shreddinglicks said:Are you saying I can factor out a e^x and leave it out as a constant?
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.
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):
##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
Considering my earlier hint, 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.
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).
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.
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.
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.
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.