How do I find the integral
#int(x*cos(5x))dx# ?

Question

Samantha Adkison · Accepted Answer

We'll remember the following integration by parts formula: #int u dv = uv - int v du# To find this integral successfully we will let #u = x#, and #dv = cos 5x dx#. Therefore, #du = dx# and #v = 1/5 sin 5x#. (#v# can be found using a quick #u#-substitution) The reason I chose #x# for the value of #u# is because I know that later on I will end up integrating #v# multiplied by #u#'s derivative. Since the derivative of #u# is just #1#, and since integrating a trig function by itself doesn't make it any more complex, we've effectively removed the #x# from the integrand and only have to worry about the sine now. Thus, entering the IBP's formula, we obtain: #int xcos5x dx = (x sin5x) / 5 - int 1/5 sin 5x dx# Pulling the #1/5# out of the integrand gives us: #int xcos5x dx = (x sin5x) / 5 - 1/5 int sin 5x dx# Integrating the sine will only take a #u#-substitution. Since we've already used #u# for the IBP's formula I'll use the letter #q# instead: #q = 5x#
#dq = 5 dx# To get a #5 dx# inside the integrand I'll multiply the integral by another #1/5#: #int xcos5x dx = (x sin5x) / 5 - 1/25 int 5sin 5x dx# And, replacing everything in terms of #q#: #int xcos5x dx = (x sin5x) / 5 - 1/25 int sinq*dq# We know that the integral of #sin# is #-cos#, so we can finish this integral off easily. Remember the constant of integration: #int xcos5x dx = (x sin5x) / 5 + 1/25 cos q + C# Now we will simply substitute back #q#: #int xcos5x dx = (x sin5x) / 5 + (cos 5x) /25 + C# Our integral is there as well.

Isabella Ackerman · Answer

undefined To find the integral of $ \int x \cos(5x) \, dx $, you can use integration by parts. Let $ u = x $ and $ dv = \cos(5x) \, dx $. Then, $ du = dx $ and $ v = \frac{1}{5} \sin(5x) $. Apply the integration by parts formula $ \int u \, dv = uv - \int v \, du $, and then substitute the values of $ u $, $ v $, $ du $, and $ dv $. You'll eventually obtain the integral in terms of $ x $ and $ \sin(5x) $.

William Abdalla · Answer

undefined To find the integral $ \int x \cos(5x) \, dx $, you can use integration by parts. Integration by parts states:

$$
\int u \, dv = uv - \int v \, du
$$

Let $ u = x $ and $ dv = \cos(5x) \, dx $. Then, differentiate $ u $ to get $ du $, and integrate $ dv $ to get $ v $:

$$
\begin{align*}
u &= x & dv &= \cos(5x) \, dx \
du &= dx & v &= \frac{1}{5} \sin(5x)
\end{align*}
$$

Apply integration by parts:

$$
\int x \cos(5x) \, dx = x \left( \frac{1}{5} \sin(5x) \right) - \int \left( \frac{1}{5} \sin(5x) \right) \, dx
$$

Now, integrate $ \frac{1}{5} \sin(5x) $ with respect to $ x $:

$$
= x \left( \frac{1}{5} \sin(5x) \right) + \frac{1}{25} \cos(5x) + C
$$

So, the integral of $ x \cos(5x) $ with respect to $ x $ is $ x \left( \frac{1}{5} \sin(5x) \right) + \frac{1}{25} \cos(5x) + C $, where $ C $ is the constant of integration.

How do I find the integral #int(x*cos(5x))dx# ?