For which of the following integrands will the table method for parts produce an antiderivative?
a) e^x tan(x)
b) x^3 sin(x)
c) e^x ln(x)
d) None of these
x^3(sin(x))
Pleaseeeee, I could understand it better if you do it because I'm so confused
To determine which of the following integrands will produce an antiderivative using the table method for parts, we need to apply the integration by parts formula. The formula is as follows:
∫ u dv = uv - ∫ v du
To use the table method for parts, we choose two functions to differentiate and integrate. We then create a table to track the derivatives and integrals of these functions. The goal is to simplify the integral by choosing functions such that the resulting integral is easier to solve.
Let's analyze each option to see if the table method for parts can be applied:
a) e^x tan(x):
To apply the table method for parts, we need two separate functions to differentiate and integrate. For e^x tan(x), we don't have two distinct functions. Therefore, we can conclude that the table method for parts cannot be directly applied to this integrand.
b) x^3 sin(x):
This integrand does have two separate functions, x^3 and sin(x), which can be used in the table method for parts. Therefore, the table method for parts can be applied to this integrand.
c) e^x ln(x):
Similar to option (a), we only have one function, e^x ln(x), and not two distinct functions to use in the table method for parts. Thus, the table method for parts cannot be directly applied here.
d) None of these:
Given the analysis above, we can see that option (d) is correct. None of the provided integrands can be solved using the table method for parts.
In summary, the correct answer is: d) None of these.
integration by parts is useful for reducing powers of x, as you can let
u = x^n, du = n x^(n-1)
apply it n times to get rid of the x factors.
The trig functions don't go away, and lnx almost always messes things up
Integration by parts is not mysterious, if you just realize that it is the product rule in disguise. You know that
(uv)' = uv' + u'v
right? Well, do the integration
∫ d(uv) = ∫ u dv + ∫ v du
Or, as it is usually presented, and since ∫ d(uv) = uv,
∫ u dv = uv - ∫ v du
So. Let's take a look at
∫x^3 sin(x) dx
u = x^3, du = 3x^2 dx
dv = sinx dx, v = -cosx
That means that
∫ u dv = uv - ∫ v du
∫ x^3 sinx dx = (x^3)(-cosx) - ∫ (-cosx)(3x^2) dx
or
-x^3 cosx + 3∫x^2 cosx dx
See that the power of x has gone from 3 to 2. That means that if you do integration by parts 2 more times, all the x stuff will be gone, leaving just cosx or sinx at the end.
Unfortunately, this does not work with
e^x tanx
because if you let
u = tanx, du = sec^2x dx, which is not simpler,
On the other hand, if you ket
dv = tanx dx, v = -ln(cosx), which is definitely not simpler.
The same complications arise with the other integrands.