integration of tan^n-2 x sec^2 x dx within limit 0 to pi/4
Let's work this one out.
d(tan(x)) / dx = d(u)/du.du/dx, u = tan(x), du/dx = sec^2(x), d(u)/du = 1
d(tan^n(x)) / dx = d(u^n)/du.du/dx, u = tan(x), du/dx = sec^2(x), d(u^n)/du = nu^(n-1)
d(tan^(n-1)(x)) / dx = d(u^(n-1))/du.du/dx, u = tan(x), du/dx = sec^2(x), d(u^(n-1))/du = (n-1)u^(n-2)
i.e. d(tan^(n-1)(x)) / dx = (n-1)u^(n-2).sec^2(x), u = tan(x)
So,
int {(n-1)tan^(n-2)(x).sec^2(x)} = tan^(n-1)(x)
And, int {tan^(n-2)(x).sec^2(x)} [0 - pi/4] = (1/(n-1)).tan^(n-1)(x) [0 - pi/4]
= (1/(n-1)).{ tan^(n-1)(pi/4) - tan^(n-1)(0) }
= (1/(n-1)).{ 1^(n-1) - 0^(n-1) }
= 1/(n-1)