More Integrals
Trig |
On each of these 2 integrals; first, I2a =
∫sec3 x dx I2a =
∫sec3 x dx =
sec x tan x - ∫sec x tan2 x dx
tan2 x = sec2x - 1 - ∫sec x tan2 x dx = - ∫sec x (sec2x - 1)dx = - ∫sec3x -
sec x)dx I2a =
∫sec3 x dx =
sec x tan x - ∫sec3x dx + ∫sec x)dx
2∫sec3 x dx =
sec x tan x + ∫sec x)dx or
∫sec3 x dx =
(sec x tan x)/2 + (ln |sec x + tanx|)/2 + C A similar strategy applies to I2b =
∫sec5 x dx In this case u =
sec3 x
dv
= sec2 x dx du
= 3 sec3 x tan x dx v =
tan x So
∫sec5 x dx =
sec3 x tan x - 3 ∫sec3 x tan2 xdx Now
3 ∫sec3 x tan2 xdx = 3 ∫sec5 x dx - 3 ∫sec3 x dx Or
∫sec5 x dx =
sec3 x tan x - 3 ∫sec5 x dx + 3 ∫sec3
x dx
4∫sec5 x dx =
sec3 x tan x + 3 ∫sec3 x dx So ∫sec5 x dx = (sec3 x tan x)/4
+ (3/4) ∫sec3 x dx |
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