P3 Revision topic - 1:15pm 9th June - OCR/Edexcel

Well the direction vector of D is AB i think

There's vector A and vector B, and you can work out the line AB. If D is a point on line AB, how do i find the position vector of D such that OD is perpendicular to AB? I don't know how to apply dot product here as I don't know anything about D...

whats tp lol

vectors are definitely my weak area

how can parametric equations be integrated and whats the best way to integrate (secx)^2 tanx dx

1) ∫sec²x.tanx using substitution u=secx

u=secx, du/dx=tanxsecx dx=du/tanxsecx therefore:

∫sec²x.tanx dx = 1) ∫sec²x.tanx du/tanxsecx, it cancels down to:
∫secx du therefore:
∫u du = (u^2)/2 + c = sec^2x+c or tanx+1

What happened to the /2 ? u=sec x so isn't ∫u du = (u^2)/2 + c = ((sec x)^2)/2 +c ?

Plus (sec x)^2 = (tan x)^2 + 1

I think you are ever so slightly confused. When you are told that u='something' you must always differentiate u wrt x. The differentition of u=sec is du/dx=tanxsecx, if you don't believe me use the quotient rule.

There's no question of u in what I wrote:
Result = (u^2)/2 + c so you replace u with its value wrt x
which gives:
((sec x)^2)/2 +c

There are no maths there


What's more you can't write: "sec^2x+c or tanx+1"
I think you meant "sec^2x+c or tan^2x+c' " (the c' is c+1, you could always just keep the term c)

Oh, yeah, I see what you meant know, I had it over two and forgot to divide, damn, thanks for pointing that out, I sometimes make silly mistakes like that.

sec^2x=tan^2x+1 therefore the answer could be:

1/2(sec^2x)+c or 1/2(tan^2x+1) +c

Final message of the day.

Good luck and stay calm, also drink lots of water but not so much so that you have to go to the toilet every 5 minutes. Don't forget your calculators!!!

Good luck everyone!!!