Question

need your help...

Answer 1

how can i prove this equation?

Answer 2

let

\[\large I_n = \int \sec^n x dx\]

Answer 3

\[\large = \int \sec^{n-2} x \sec^2 x dx\]

Answer 4

integration by parts :
\(u = \sec^{n-2}x\)
\(v = \tan x \)

Answer 5

\[\large = \sec^{n-2}x \tan x - \int \left( \sec^{n-2} x\right)' \tan x dx\]

Answer 6

\[\large = \sec^{n-2}x \tan x - \int (n-2)\left( \sec^{n-3} x \right)\left(\sec x \tan x \right) \tan x dx\]

Answer 7

\[\large = \sec^{n-2}x \tan x - (n-2) \int \sec^{n-2} \tan^2 x dx\]

Answer 8

\[\large = \sec^{n-2}x \tan x - (n-2) \int \sec^{n-2} \left( \sec^2 x -1\right) dx\]

Answer 9

\[\large = \sec^{n-2}x \tan x - (n-2) \int \sec^n x dx + (n-2) \int \sec^{n-2}x ~dx\]

Answer 10

\[\large = \sec^{n-2}x \tan x - (n-2) I_n + (n-2) \int \sec^{n-2}x ~dx\]

Answer 11

remember that there is \(I_n\) on the left hand side :

\[\large I_n = \sec^{n-2}x \tan x - (n-2) I_n + (n-2) \int \sec^{n-2}x ~dx\]

Answer 12

send the right hand side \(I_n\) term to left side, and isolate \(I_n\)

Answer 13

\[\large I_n + (n-2) I_n = \sec^{n-2}x \tan x + (n-2) \int \sec^{n-2}x ~dx\]

\[\large I_n (n-1) = \sec^{n-2}x \tan x + (n-2) \int \sec^{n-2}x ~dx\]

Answer 14

divide n-1 both sides

Answer 15

wow~amazing!!

Answer 16

thanks your help!

Answer 17

np :)