Hard Integration Question (1 Viewer)

[hyparzero]

Can someone please enlighten me on how to actually do this question:

Q1. Find ∫(sin^max)/(cosⁿax) in terms of m, n, Cos(ax) and sin(ax)

Help would be appreciated!

 

[Yip]

Let p=(sin^m-1ax)/(cos^n-1ax)

dp/dx=[[a(m-1)(sin^m-2ax)(cosax)(cos^n-1ax)]-[a(n-1)(cos^n-2ax)(-sinx)(sin^m-1ax)]]/(cos^2n-2ax)
=[a(m-1)(sin^m-2ax)(cosⁿax)+a(n-1)(cos^n-2ax)(sin^max)]/(cos^2n-2ax)
=[a(m-1)(sin^m-2ax)/(cos^n-2ax)]+a(n-1)(sin^max)]/(cosⁿax)

Let I[m,n subscript]=∫(sin^max)/(cosⁿax)

Integrating the expression for dp/dx w.r.t.x,

p=(sin^m-1ax)/(cos^n-1ax)=a(m-1)I[m-2,n-2]+a(n-1)I[m,n]
I[m,n]=[1/a(n-1)][[(sin^m-1ax)/(cos^n-1ax)]-a(m-1)I[m-2,n-2]]

I think thats right............if u really want it fully in terms of sin(ax) and cos(ax), you can keep repeatedly applying the reduction formula, but it looks a little messy to me.....