Can Integration by Parts Solve Higher Power Trigonometric Integrals?

[just_see]

Homework Statement

If \int_{0}^{pi/4}\tan^{6}(x)sec(x)dx = I then express \int_{0}^{pi/4}\tan^{8}(x)sec(x)dx in terms of I.

Homework Equations

The Attempt at a Solution

I figured you would have to get the second equation to look like the first one, so I pulled out a tan^2(x) to make it
\int_{0}^{pi/4}\((sec^{2}(x)-1)tan^{6}(x)sec(x)dx then
\int_{0}^{pi/4}\sec^{2}{x}tan^{6}(x)sec(x)dx - I

And here's where I get stuck. My thought is to use integration by parts, setting sec^2 = u, and tan^6xsecxdx = dv. However this just leads me to a dead end when I try to figure out what v could be. Using the opposite substitution (u = tan^6... etc) I end up with

tan^{7}(x)sec(x) - 6\int_{0}^{pi/4}\tan^{6}(x)sec^{3}(x)dx = 2\int_{0}^{pi/4}\tan^{8}(x)sec(x)dx

which gives me the same problem I had to begin with (an even power to tangent and an odd power of secant).

Is there some substitution I could make somewhere?

 

[TMFKAN64]

Have you tried integrating by parts using dv = sec x tan x?

 

[just_see]

Yeah... that gets me:

\int_{0}^{pi/4}\sec^{2}{x}tan^{6}(x)sec(x)dx - I

u = sec^2xtan^5x dv = secxtanx
du = 2sec^2xtan^6x + 5tan^4xsec^4x

sec^{3}(x)tan^{5}(x) - \int_{0}^{pi/4}\sec(x)[2sec^{2}(x)tan^{6}(x) + 5tan^{4}(x)sec^{4}(x)]dx

simplified...

sec^{3}(x)tan^{5}(x) - 2\int_{0}^{pi/4}\sec^{3}(x)tan^{6}(x) - 5\int_{0}^{pi/4}\tan^{4}(x)sec^{5}(x)dx

Which still leaves me with odd powers of secant and even powers of tangent...

EDIT: Woops, dropped the "- I" after all of those, but it's pretty irrelevant.

 

[just_see]

Hmm.. I just tried a different approach, would like to know if this is a correct approach...

\int_{0}^{pi/4}\sec^{2}{x}tan^{6}(x)sec(x)dx - I

using u = secx, dv = tan^6xsec^2x
du = secxtanx, v = (1/7)tan^7x

\frac{sec(x)tan^{7}(x)}{7} - \frac{1}{7}\int_{0}^{pi/4}\sec(x)tan^{8}(x)dx - I = \int_{0}^{pi/4}tan^{8}(x)sec(x)dx

So...

\frac{\sqrt{2}}{7} - I = \frac{8}{7}\int_{0}^{pi/4}tan^{8}(x)sec(x)dx

= \frac{\sqrt{2}-7I}{8}this look right?

 

[TMFKAN64]

Looks good to me. (And for the record, you can get the same result by using using dv = sec x tan x in the *original* equation. You'll get an integral with sec^3 x, which you can rewrite as sec x(1 + tan^2 x).)

 

[just_see]

Looks good to me. (And for the record, you can get the same result by using using dv = sec x tan x in the *original* equation. You'll get an integral with sec^3 x, which you can rewrite as sec x(1 + tan^2 x).)

Ah... yeah I see. I did that at one point but didn't think to write sec^2 as (1+tan^2). Anyway, thanks a ton.