Tangent Integral!

Calculus Level 3

Let ${ U }_{ n } = \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ n }{ x }}dx$

Where $n\in Z$ with $n>1$

Express ${ U }_{ n }+{ U }_{ n-2 }$ in terms of n

Hence solve ${ U }_{ 6 }$

${ U }_{ 6 }$ can be represented as $\frac { a }{ b } -\frac { \pi }{ c }$ for positive integers $a,b,c$ with $a,b$ coprime.

Find $a+b+c$ .

1 solution

Will McGlaughlin
May 6, 2018

${ U }_{ n }+{ U }_{ n-2 } = \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ n }{ x }}dx + \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ n-2 }{ x }}dx$

$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ n }{ x } (1+\cot ^{ 2 }{ x } )dx }$

$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ n }{ x } \csc ^{ 2 }{ x } dx }$

$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ n-2 }{ x } \tan ^{ 2 }{ x } \csc ^{ 2 }{ x } dx }$

$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan ^{ n-2 }{ x } \sec ^{ 2 }{ x } dx }$

$u=tan(x),\quad du=\sec ^{ 2 }{ x } dx$

$x=0,\quad u=0,\quad x=\frac { \pi }{ 4 } ,\quad u=1$

$\int _{ 0 }^{ 1 }{ { u }^{ n-2 }dx } =\frac { 1 }{ n-1 }$

${ U }_{ 6 }+{ U }_{ 4 }=\frac { 1 }{ 6-1 }$

${ U }_{ 6 }=\frac { 1 }{ 6-1 } -{ U }_{ 4 }$

${ U }_{ 4 }+{ U }_{ 2 }=\frac { 1 }{ 4-1 }$

${ U }_{ 4 }=\frac { 1 }{ 4-1 } -{ U }_{ 2 }$

${ U }_{ 2 }=\frac { 1 }{ 2-1 } -{ U }_{ 0 }$

${ U }_{ 0 }=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ dx }$

${ U }_{ 0 }=\frac { \pi }{ 4 }$

${ U }_{ 6 }=\frac { 1 }{ 5 } -\frac { 1 }{ 3 } +1-\frac { \pi }{ 4 }$

${ U }_{ 6 }=\frac { 13 }{ 15 } -\frac { \pi }{ 4 }$

Therefore a + b + c = $\boxed{32}$