Use integration by parts to evaluate the definite integral.

Intergral: xsec^2(6x) dx

I set u=sec^2(6x) and dv=x, but the problems seems to get harder as I go on.

switch it around

let u = x
du/dx = 1
du = dx

let dv = (sec (6c))^2 dx
v = (1/6)tan(6x)

I recalled that d(tanx)/dx = sec^2 x

the we have
uv - [integral]v du
= (1/6)(sec(6x))^2(tan6x) - [int.](1/6)tan(6x) dx

but I recall that d(ln(cos))/dx
= -sinx/cosx = -tanx
so we can integrate tan(6x)

I will leave the "cleanup" for you, but please check my steps.
I did not write them down on paper, and I have a tendency lately to make typing errors.

If sec^2 (6x) dx = dv, v = (1/6) tan 6x

u = x and du = dx

uv - Integral v du
= (x/6) tan(6x) - (1/6) integral tan (6x) dx
= (x/6) tan(6x) + (1/36) log cos(6x)

thanks drwls for catching my error, looks like I messed up in the uv part

I didn't see Reiny's earlier answer when I started on it. I was afraid my answer might be wrong because I was in a hurry and didn't double check it. Jake, I suggest you verify it by differentiating my answer.

I used an integral table for the sec^2 6x and tan(6x) integrals. Life is too short to do all that stuff.

When using integration by parts, the goal is to choose which part of the integral to differentiate and which part to integrate. In this case, you correctly identified that u should be chosen as sec^2(6x) and dv as x.

Now, let's proceed with the integration by parts formula:

∫ u dv = uv - ∫ v du

Taking the differentials of u and v, we have:
du = 2sec(6x)tan(6x)dx
dv = xdx

Let's substitute the values into the formula:

∫ xsec^2(6x) dx = ∫ u dv = uv - ∫ v du

To find uv, we multiply u and v together:
uv = sec^2(6x) * (1/2)x^2 = (1/2)x^2 sec^2(6x)

Now let's find ∫ v du:
∫ v du = ∫ x * du = ∫ x * 2sec(6x)tan(6x)dx

To solve this integral, a trigonometric substitution can be used:
Let's substitute u = 6x, so that du = 6dx.

After substitution, the integral becomes:
∫ x * 2sec(6x)tan(6x)dx = ∫ (1/3)(u/6) * 2sec(u)tan(u) * (du/6)

Simplifying, we have:
(1/9) ∫ u sec(u)tan(u) du

At this point, we can use integration techniques such as integration by parts again or trigonometric identities, but the resulting integral may become more complicated.

Therefore, it would be better to evaluate the original integral using another method, such as a substitution.