Use integration by parts to evaluate the definite integral.

Question

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.

Answer 1

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.

Answer 2

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)

Answer 3

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

Answer 4

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.