Question

Integrals,


Can you please help me outo n this thingy? http://screencast.com/t/PiDhvWM2i

Answer 1

For the first one, maybe use: \[
u=\sec^{n-2}(x)\quad dv=\sec^2(x)dx
\]

Answer 2

I dunno, I just want \(dv\) to be easy to integrate.

Answer 3

Yah that seems like a good way to start :)

Answer 4

hmm

Answer 5

Split into \(\sec^{n-2}x\sec^2x=\sec^{n-2}(\tan^2 x+1)\)

Answer 6

\[\Large \int\limits \sec^{n-2}x\;\sec^2x\;dx\]By parts the first time gives us:\[\Large \sec^{n-2}x\;\tan x-(n-2)\int\limits \sec^{n-3}x(\sec x \tan x) \tan x\;dx\]

Answer 7

I almost got it, only thing left is for me to understand this exact step http://screencast.com/t/TLQRq9ZF9

Answer 8

There, more clear picture of those 2 steps: http://screencast.com/t/5fbV6IhfpEYd

Answer 9

Give your original integral a variable name, its easier to keep track of that way.\[\Large I_n\quad=\quad \int\limits \sec^n x\;dx\]So it looks like you've gotten it to the point of:\[\Large I_n\quad=\quad \sec^{n-2}x\;\tan x-(n-2)I_n+(n-2)I_{n-2}\]

Answer 10

Understand how I rewrote the right side?
We want to get all of the \(\Large I_n\) to the left side.

Answer 11

Oh yea, got it ! Thanks everything is good ... for now !! :P

Answer 12

cool c: