(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

[tex]\int_{0}^{pi/4}\((sec^{2}(x)-1)tan^{6}(x)sec(x)dx[/tex] then

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

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

[tex]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[/tex]

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Trig Integral Problem

**Physics Forums | Science Articles, Homework Help, Discussion**