Question-96693

Question Number 96693 by 175 last updated on 03/Jun/20

Answered by abdomathmax last updated on 04/Jun/20

A_{n} = \int \frac{dx}{1 + \tan^{n} x} =_{tanx = t} \int \frac{dt}{(1 + t^{2}) (1 + t^{n})}

let decompose F (t) = \frac{1}{(t^{2} + 1) (t^{n} + 1)} = \frac{1}{(t - i (t + i) (t^{n} + 1)}

\Rightarrow F (t) = \frac{a}{t - i} + \frac{b}{t + i} + \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}}

z_{p} roots of t^{n} + 1 = 0 \Rightarrow z_{p} = e^{i \frac{(2 k + 1) π}{p}} p \in [[0, n - 1]]

a = \frac{1}{2 i (1 + i^{n})} and b = \frac{- 1}{2 i (1 + {(- i)}^{n})} \Rightarrow

F (t) = \frac{1}{2 i (i^{n} + 1) (t - i)} - \frac{1}{2 i (1 + {(- i)}^{n}) (t + i)} + \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}}

= \frac{1}{2 i} {\frac{(1 + {(- i)}^{n}) (t + i) - (1 + i^{n}) (t - i)}{(1 + i^{n}) (1 + {(- 1)}^{n}) (t^{2} + 1)}}

+ \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}}

= \frac{α t + β}{t^{2} + 1} + \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}} \Rightarrow

\frac{1}{(t^{2} + 1) (t^{n} + 1)} - \frac{α t + β}{t^{2} + 1} = \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}} \Rightarrow

\frac{1}{(t^{2} + 1)} {\frac{1}{t^{n} + 1} - (α t + β)} = Σ (\dots) \Rightarrow

\frac{1 - (α t + β) (t^{n} + 1)}{(t^{2} + 1) (t^{n} + 1)} = \sum_{p = 0}^{\infty} \frac{a_{p}}{t - z_{p}}

a_{p} = \frac{f (z_{p)}}{g^{^{'}} (z_{p})} we have f (z_{p}) = \frac{1 - (α z_{p} + β) (z_{p}^{n} + 1)}{(z_{p}^{2} + 1) (z_{p}^{n} + 1)}

g (t) = t^{n + 2} + t^{2} + t^{n} + 1 \Rightarrow

g^{^{'}} (t) = (n + 2) t^{n + 1} + {nt}^{n - 1} + 2 t \Rightarrow

g^{^{'}} (z_{p}) = (n + 2) z_{p}^{n + 1} + {nz}_{p}^{n - 1} + {2 z}_{p} \Rightarrow

\int F (t) dt = aln (t - i) + bln (t + i) + \sum_{p = 0}^{n - 1} a_{p} \ln (t - z_{p}) + C

Commented by 175 last updated on 04/Jun/20

nice

Commented by mathmax by abdo last updated on 04/Jun/20

thankx