Integrating tan^5(6x) sec^3(6x) - A Guide

  • Thread starter Thread starter grothem
  • Start date Start date
  • Tags Tags
    Integrating
grothem
Messages
21
Reaction score
0

Homework Statement


\int tan^5(6x) sec^3(6x) dx



Homework Equations





The Attempt at a Solution


first off I set u=6x to get 1/6\int tan^5(u) sec^3(u) dx
then I used trig identities to put tangent in terms of secant and I came up with

\int sec^9(u)-3sec^7(u)+3sec^5(u)-sec^3(u) dx
Not sure where to go from here, or if I'm doing this the right way
 
Physics news on Phys.org
Watch your substitutions.

\frac 1 6\int\tan^{5}u\sec^{3}udu

\frac 1 6\int\tan^{4}u\sec^{2}\sec u\tan udu

*\tan^{2}u+1=\sec^{2}u

Take it from here.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top