integration problem

balzarywex

does anyone have any tips for for finding the integral of (sec x)^4 dx?
iv looked thorugh the table of integrals but no inspiration as of yet! any help would be greatly appreciated!

JC 2K3

hmmm

try writing it as ((sec x)^2)((sec x)^2) dx and do integration by parts.

(integ of (sec x)^2 dx = tanx + c)

Useless Fecker

Is the answer 2?

Kevster

Sec(x) is 1/Cos(x)

Therefore, Sec(x)^4 is 1/Cos(x)^4

1/Cos(x)^4 is then Cos(x)^-4

I think you can proceed much easier from that point.

...although, JC 2K3's way is obviously much much simpler!

JC 2K3

((sec x)^2)((sec x)^2) dx

u = sec(x)^2 = cos(x)^-2
du = -2cos(x)^-3(-sinx).dx = 2sin(x)/cos(x)^3.dx = 2tan(x)/cos(x)^2.dx
dv=sec(x)^2.dx
v = tan(x)

uv = tan(x)/cos(x)^2 = tan(x)sec(x)^2

integ(v.du) = 2(integ(tan(x)^2/cos(x)^2.dx))
= 2(integ((tan(x)^2)(sec(x)^2).dx))

w = tan(x)^2
dw = 2sec(x)^2(tan(x))dx
2(integ((tan(x)^2)(sec(x)^2).dx)) = integ(w^(1/2).dw)
=2w^(3/2)/3
2tan(x)^3/3

uv - integ(v.du) = tan(x)/cos(x)^2 - 2tan(x)^3/3
tan(x)sec(x)^2 - 2tan(x)^3/3
(3tan(x)sec(x)^2 - 2tan(x)^3)/3
(1/3)(3tan(x)sec(x)^2 - 2tan(x)^3)


Hopefully you can follow that, I think it's right. I'm only doing my LC, so perhaps there's a quicker way to do it, I dunno, but hope I helped.

Kevster

Phwoarrr..... I need a longer sheet with the way I'm doing it. I'm managing to do nested 'Integral by parts' if you get me. So far, I've nested it thrice.

...make that fourth. I think I'll give-up.

Crumbs

That integration by parts method looks right but I think there's a quicker way.

Use the identity: sec^2(x) = 1 + tan^2(x)

Int [sec^4(x) dx] = Int [(1 + tan^2(x)).sec^2(x) dx]

Substitution:
u = tan x
du = sec^2(x).dx

Int [1 + u^2 du]

= u + (u^3)/3 + c

= tan x + (tan^3(x))/3 + c

JC 2K3

Ah, damn you!

If you replace the sec(x)^2 with (1+tan(x)^2) in my answer you get the same answer:

(1/3)(3tan(x)sec(x)^2 - 2tan(x)^3)
= (1/3)(3tan(x)(1+tan(x)^2) - 2tan(x)^3)
= (1/3)(3tan(x) + 3tan(x)^3 - 2tan(x)^3)
= (1/3)(3tan(x) + tan(x)^3)
= tan(x) + (tan(x)^3)/3

balzarywex

you boardsies are legends for help! too bad i forgot i posted up here and got the answer two days ago (the 1/Cosx way!) cheers!