Limit and Integral on tan

$\begin{aligned} t(n) & = \int_0^\frac \pi 4 \tan^{2n} x \ dx \\ & = \int_0^\frac \pi 4 \tan^2 x \tan^{2(n-1)} x \ dx \\ & = \int_0^\frac \pi 4 (\sec^2 x -1 )\tan^{2(n-1)} x \ dx \\ & = {\color{#3D99F6} \int_0^\frac \pi 4 \sec^2 x \tan^{2(n-1)} x \ dx} - t(n-1) & \small \color{#3D99F6} \text{Let }u = \tan x \implies du = \sec^2 x \ dx \\ & = {\color{#3D99F6} \int_0^1 u^{2(n-1)}\ du} - t(n-1) \\ & = \frac 1{2n-1} - t(n-1) \end{aligned}$

Now we have:

$\begin{aligned} t(1) & = \frac 1{2(1)-1} - \int_0^\frac \pi 4 dx = 1 - \frac \pi 4 \\ \implies t(n) & = \sum_{k=1}^n \frac {(-1)^{n+k}}{2n-1} + (-1)^n\frac \pi 4 \\ t(5) & = \frac 19 - \frac 17 + \frac 15 - \frac 13 + 1 - \frac \pi 4 = \frac {263}{315} - \frac \pi 4 \end{aligned}$

Therefore, $b-a = 315-263 = \boxed{52}$ .

Let,

$f(x)=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ x+2 }{ t } +\tan ^{ x }{ t } dt }$

$f(x)=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ x }{ t } (\tan ^{ 2 }{ t } +1)dt }$

$f(x)=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ x }{ t } (\tan ^{ 2 }{ t } +1)dt }$

Let $v=tan(t)$

$d(tan(t))=d(v)$

$(\tan ^{ 2 }{ t } +1)d(t)=d(v)$

$f(x)=\int _{ 0 }^{ 1 }{ { v }^{ x }dv }$

$f(x)=\frac { 1 }{ x+1 }$

$t(5)=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ 10 }{ x } dx }$

$t(5)=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ 10 }{ x } +\tan ^{ 8 }{ x } -\tan ^{ 8 }{ x } -\tan ^{ 6 }{ x } +\tan ^{ 6 }{ x } +\tan ^{ 4 }{ x } -\tan ^{ 4 }{ x } -\tan ^{ 2 }{ x } +\tan ^{ 2 }{ x+\tan ^{ 0 }{ x } } -1\quad dx }$

$t(5)=f(8)-f(6)+f(4)-f(2)+f(0)-\int _{ 0 }^{ \frac { \pi }{ 4 } }{ dx }$

$t(5)=f(8)-f(6)+f(4)-f(2)+f(0)-\frac { \pi }{ 4 }$

$t(5)=\frac { 1 }{ 9 } -\frac { 1 }{ 7 } +\frac { 1 }{ 5 } -\frac { 1 }{ 3 } +\frac { 1 }{ 1 } -\frac { \pi }{ 4 }$

$t(5)=\frac { 263 }{ 315 } -\frac { \pi }{ 4 }$

$\frac { a }{ b } =\frac { 263 }{ 315 }$

$b-a=52$