SQT thread

So what exactly does this coefficient represent?

I have absolutely no clue what you're talking about, and that still doesn't explain why you can use IBP without two things being multiplied together.

Mentally tried normal substituion, back substituion and by parts and there is no way.

Trig sub unless you have some genius identity.

Hmm, and trigonometric substituation produces a gigantic mess.
Which leads me to suspect that I'm doing this question wrong.
I assumed the answer was to integrate the function for the distance of a part of the rod from P.

Why hasn't GR been superseded yet if it doesn't fit the Standard Model in quantum physics with gravity and with other phenomena like dark energy?

He is right.

You are multiplying by dx. You always multiply by dx.

>that still doesn't explain why you can use IBP without two things being multiplied together.
Uh.... yes it does. Did you actually read what he posted?

If I could understand it when it's written in such a shorthand way, I wouldn't be posting in a SQT.

The two things you are multiplying are [math]\arcsin(3x)[/math] and [math]dx[/math].

Let [math]u=\arcsin(3x)[/math] and [math]dv=dx[math].
Then [math]du=\frac{3}{\sqrt{1-9x^2}}dx[/math] and [math]v=x[/math] (since the antiderivative of 1 is x).

So we have...
[math]\int\arcsin(3x)dx=x\arcsin(3x)-\int\frac{3x\,dx}{\sqrt{1-9x^2}}[/math]
by integration by parts.

[math]\int\arcsin(3x)dx=x\arcsin(3x)-\int\frac{3x\,dx}{\sqrt{1-9x^2}}[/math]

jfc