Integral Help!

[I Love Math]

Hello! I've had a little roadblock in my homework and I'm hoping someone here might be able to help. I'm trying to take the antiderivative of secx^3 (secant cubed).

I replaced secx^2 with 1+tanx^2 and then distributed to get secx+secxtanx^2. Now I'm stuck with secxtanx^2. I've tried integration by parts and I end up going in circles. I know I'm probably missing something simple. I work in a college math lab and I've asked the other tutors, but they are as clueless as I am on this one.

I've checked in the book and it says if the power of tangent is even and the power of secant is odd, there is no standard method of evaluation and to possibly use integration by parts. The other Calculus book I have recommends I use tables, but that's definitely not a good idea when I am going to need to know how to *get* the answer and not how to look it up.

Thanks in advance for any insight!

 

[soroban]

Hello, I Love Math!

This is a classic integration problem . . . There's nothing quite like it!
. . I love this procedure . . . I'll baby-step through it for you.

I'm trying to take the antiderivative of sec³x.

I replaced sec²x with 1 + tan²x

and then distributed to get sec x + sec x tan²x

Now I'm stuck with sec x tan²x.

I've tried integration by parts and I end up going in circles.
There's a reason for that . . . you're supposed to!

Let .I . = . ∫ sec³x dx

As you said: . sec³x .= .sec x(sec²x) .= .sec x(1 + tan²x) .= .sec x + sec x tan²x

So we have: . I . = . ∫ sec x dx .+ .∫ sec x tan²x dx

We do the second integral by-parts:
. . . u = tan x . . . . . . dv = sec x tan x dx
. . .du = sec²x dx . . . . v = sec x

We have: . I . = . ln|sec x + tan x| .+ .sec x tan x .- .∫ sec³x dx
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \______/ . This is I !

So we have: . I . = . sec x tan x + ln|sec x + tan x| - I + C

. . . . .Then: . 2I . = . sec x tan x + ln|sec x + tan x| + C

. .Therefore: . I . = . (1/2)[sec x tan x + ln|sec x + tan x|] + C . . . There!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This has always struck me as amusing . . .

We want the integral of secant-cubed.
We went through all those strange steps and came up with The Answer.

Yet (think about it) we never integrated secant-cubed!
. . . We did "everything but".

Am I the only one laughing?

 

[tkhunny]

I got lost in the details when I looked at it briefly, earlier, but the irony was not lost.

 

[I Love Math]

Thanks so much! I got there, but I integrated by parts again and everything cancelled out. My professor always says "Anything that's worth doing is worth doing right and anything that's worth doing right is worth doing *wrong* first." I'll never get stuck on this integral again!