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
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
4 answers
Pleaseeeee, I could understand it better if you do it because I'm so confused
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.
(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.
x^3(sin(x))
1 answer