Finding the integral of (sec x)^3 dx using integration by parts

[CinderBlockFist]

Hi all, I am stuck on this trig integral problem. The answer is provided in the book, but I do not know how to get it. The problem is this:


Integral of (secX)^3 dx


it says first use integration by parts:

u = sec x du = sec x tan x dx

dv = (secx)^2 dx v = tan x

uv - integral (v du) = sex x tan x - integral sec x((secx)^2-1) dx


Then it says secx tanx - integral (sec x)^3 dx + integral sec x dx


Now there is another (sec x)^3 like in the original problem, after this step they just provide the answer, but how ?




The answer is 1/2(secxtanx + ln |secx+tanx|) + C

 

[arildno]

Let
$$I=\int\frac{1}{\cos^{3}x}dx$$
Hence, you have shown:
$$I=sec(x)tan(x)-I+\int\frac{1}{\cos{x}}dx$$
Can you take it from there?

 

[CinderBlockFist]

hmm, let me try, that's diff. method from the book but it looks easier, brb. thx.