if-I-n-0-pi-4-tan-n-xdx-and-I-n-a-n-b-n-I-n-2-then-find-a-10-

Question Number 78458 by kaivan.ahmadi last updated on 17/Jan/20

i f I_{n} = \int_{0}^{\frac{π}{4}} {t a n}^{n} x d x a n d I_{n} = a_{n} + b_{n} I_{n - 2} t h e n f i n d a_{10}

Commented by kaivan.ahmadi last updated on 17/Jan/20

I_{n} + I_{n - 2} = \int {t g}^{n} x d x + \int {t g}^{n - 2} x d x = \int ({t g}^{n} x d x + {t g}^{n - 2} x) d x

= \int {t g}^{n - 2} x ({t g}^{2} x + 1) d x

u = t g x \Rightarrow d u = (1 + {t g}^{2} x) d x \Rightarrow \int u^{n - 2} d u =

\frac{u^{n - 3}}{n - 1} = \frac{{t g}^{n - 3} x}{n - 1} ∣_{0}^{\frac{π}{4}} = \frac{1}{n - 1}

\Rightarrow

I_{n} + I_{n - 2} = \frac{1}{n - 1}

\Rightarrow I_{n} = \frac{1}{n - 1} - I_{n - 2} = a_{n} + (- 1) I_{n - 2} \Rightarrow a_{n} = \frac{1}{n - 1}

\Rightarrow a_{10} = \frac{1}{9}

Commented by mathmax by abdo last updated on 17/Jan/20

I_{n} = \int_{0}^{\frac{π}{4}} {t a n}^{n} x d x \Rightarrow I_{n} = \int_{0}^{\frac{π}{4}} {t a n}^{n - 2} x (1 + {t a n}^{2} x - 1) d x

= \int_{0}^{\frac{π}{4}} (1 + {t a n}^{2} x) {t a n}^{n - 2} x d x - \int_{0}^{\frac{π}{4}} {t a n}^{n - 2} x d x

= {[\frac{1}{n - 1} {t a n}^{n - 1} x]}_{0}^{\frac{π}{4}} - I_{n - 2} = \frac{1}{n - 1} - I_{n - 2} i f n ⩾ 2 \Rightarrow a_{n} = \frac{1}{n - 1}

a n d b_{n} = - 1 \Rightarrow a_{10} = \frac{1}{9}

l e t c a l c u l a t e I_{n} i n t e r m s o f n w e h a v e

I_{n} + I_{n - 2} = \frac{1}{n - 1} \Rightarrow I_{2 n} + I_{2 n - 2} = \frac{1}{2 n - 1} l e t I_{2 n} = u_{n} \Rightarrow

u_{n} + u_{n - 1} = \frac{1}{2 n - 1} \Rightarrow \sum_{k = 1}^{n} {(- 1)}^{k} (u_{k} + u_{k - 1}) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} \Rightarrow

- (u_{1} + u_{0}) + u_{2} + u_{1} + \dots . {(- 1)}^{n} (u_{n} + u_{n - 1}) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1}

{(- 1)}^{n} u_{n} - u_{0} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} \Rightarrow {(- 1)}^{n} u_{n} = u_{0} + \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} \Rightarrow

u_{n} = {(- 1)}^{n} u_{0} + {(- 1)}^{n} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} = \frac{π}{4} {(- 1)}^{n} + {(- 1)}^{n} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1}

\Rightarrow I_{2 n} = \frac{π}{4} {(- 1)}^{n} + {(- 1)}^{n} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k - 1} l e t d e t e m i n e I_{2 n + 1} ?

w e h a v e I_{2 n + 1} + I_{2 n - 1} = \frac{1}{2 n} \Rightarrow \sum_{k = 1}^{n} {(- 1)}^{k} (v_{k} + v_{k - 1}) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k}

(v_{n} = I_{2 n + 1}) \Rightarrow {(- 1)}^{n} v_{n} - v_{0} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k} \Rightarrow

{(- 1)}^{n} v_{n} = v_{0} + \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k} \Rightarrow

v_{n} = {(- 1)}^{n} v_{0} + {(- 1)}^{n} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k}

{v_{0} = I_{1} = \int_{0}^{\frac{π}{4}} t a n x d x = - \int_{0}^{\frac{π}{4}} \frac{- s i n x}{c o s x} d x = - l n ∣ c o s x ∣]}_{0}^{\frac{π}{4}}

= - l n (\frac{1}{\sqrt{2}}) = \frac{1}{2} l n (2) \Rightarrow I_{2 n + 1} = {(- 1)}^{n} \frac{l n (2)}{2} + {(- 1)}^{n} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{2 k}