integration of tan^n-2 x sec^2 x dx within limit 0 to pi/4

Question

solve the problem step by step. which integral formula should I use and how

Answer 1

Let u=tan^p(x); du=ptan^(p-1)(x)sec^2(x)dx.

Therefore, if pJ=∫ptan^(p-1)(x)sec^2(x)dx=tan^p(x), J=tan^p(x)/p.

So If we now put p-1=n-2, then p=n-1 and J=tan^(n-1)(x)/(n-1).

For {0,π/4}, J=(tanπ/4)^(n-1)/(n-1)=1/(n-1).

Answer 2

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)