Harder Integration

∫dx/(asinx + bcosx)^n = ∫d(x-u)/(Kcos(x-u))^n
where K^2=a^2+b^2, tan u = a/b, and
∫d(x-u)/(Kcos(x-u))^n
=K^(-n) ∫sec^n (x-u) d(x-u)
=K^(-n) ∫sec^(n-2)(x-u) d tan(x-u) (suppose n≧2)
=K^(-n) tan(x-u)sec^(n-2) (x-u) -K^(-n) ∫(n-2)sec^(n-2) (x-u) tan^2 (x-u)dx
=K^(-n) tan(x-u)sec^(n-2) (x-u) -(n-2)K^(-n) ∫sec^(n-2)(x-u) (sec^2 (x-u)-1)dx
=K^(-n) tan(x-u)sec^(n-2) (x-u) -(n-2)∫dx/(Kcos(x-u))^n + (n-2)/K^2∫dx/(Kcos(x-u))^(n-2)

Therefore
(n-1)∫d(x-u)/(Kcos(x-u))^n
=(1/K)sin(x-u)/(Kcos(x-u))^(n-1)+(n-2)/K^2 ∫dx/(Kcos(x-u))^(n-2)

∫dx/(asinx + bcosx)^n
=(1/(n-1)K)(sinxcosu-cosxsinu)/(asinx+bcosx)^(n-1)
+ (n-2)/(n-1)K^2∫dx/(asinx + bcosx)^(n-2)
=(Asinx+Bcosx)/(asinx+bcosx)^(n-1) + C∫dx/(asinx + bcosx)^(n-2)

where A=cosu/K(n-1), B=-sinu/K(n-1), C=(n-2)/(n-1)K^2

PS. I think the last (n-1) should be (n-2)