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 x dx
Now 3 ∫ sec3 x tan2
x dx
= 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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