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

	Answered by abdomathmax last updated on 04/Jun/20
	$A_n =∫ (dx/(1+tan^n x)) =_(tanx =t) ∫ (dt/((1+t^2 )(1+t^n ))) let decompose F(t) =(1/((t^2 +1)(t^n +1))) =(1/((t−i(t+i)(t^n +1))) ⇒F(t) =(a/(t−i)) +(b/(t+i)) +Σ_(p=0) ^(n−1) (a_p /(t−z_p )) z_p roots of t^n +1=0 ⇒z_p =e^(i(((2k+1)π)/p)) p ∈[[0,n−1]] a =(1/(2i(1+i^n ))) and b =((−1)/(2i(1+(−i)^n ))) ⇒ F(t)=(1/(2i(i^n +1)(t−i))) −(1/(2i(1+(−i)^n )(t+i))) +Σ_(p=0) ^(n−1) (a_p /(t−z_p )) =(1/(2i)){(((1+(−i)^n )(t+i)−(1+i^n )(t−i))/((1+i^n )(1+(−1)^n )(t^2 +1)))} +Σ_(p=0) ^(n−1) (a_p /(t−z_p )) =((αt +β)/(t^2 +1)) +Σ_(p=0) ^(n−1) (a_p /(t−z_p )) ⇒ (1/((t^2 +1)(t^n +1)))−((αt+β)/(t^2 +1)) =Σ_(p=0) ^(n−1) (a_p /(t−z_p )) ⇒ (1/((t^2 +1))){(1/(t^n +1))−(αt +β)} =Σ(...) ⇒ ((1−(αt +β)(t^n +1))/((t^2 +1)(t^n +1))) =Σ_(p=0) ^∞ (a_p /(t−z_p )) a_p =((f(z_(p)) )/(g^′ (z_p ))) we have f(z_p ) =((1−(αz_p +β)(z_p ^n +1))/((z_p ^2 +1)(z_p ^n +1))) g(t) =t^(n+2) +t^2 +t^n +1 ⇒ g^′ (t) =(n+2)t^(n+1) +nt^(n−1) +2t ⇒ g^′ (z_p ) =(n+2)z_p ^(n+1) +nz_p ^(n−1) +2z_p ⇒ ∫ F(t)dt =aln(t−i)+bln(t+i)+Σ_(p=0) ^(n−1) a_p ln(t−z_p ) +C$
	$A_{n} = \int \frac{dx}{1 + \tan^{n} x} =_{tanx = t} \int \frac{dt}{(1 + t^{2}) (1 + t^{n})}$ $let decompose F (t) = \frac{1}{(t^{2} + 1) (t^{n} + 1)} = \frac{1}{(t - i (t + i) (t^{n} + 1)}$ $\Rightarrow F (t) = \frac{a}{t - i} + \frac{b}{t + i} + \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}}$ $z_{p} roots of t^{n} + 1 = 0 \Rightarrow z_{p} = e^{i \frac{(2 k + 1) π}{p}} p \in [[0, n - 1]]$ $a = \frac{1}{2 i (1 + i^{n})} and b = \frac{- 1}{2 i (1 + {(- i)}^{n})} \Rightarrow$ $F (t) = \frac{1}{2 i (i^{n} + 1) (t - i)} - \frac{1}{2 i (1 + {(- i)}^{n}) (t + i)} + \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}}$ $= \frac{1}{2 i} {\frac{(1 + {(- i)}^{n}) (t + i) - (1 + i^{n}) (t - i)}{(1 + i^{n}) (1 + {(- 1)}^{n}) (t^{2} + 1)}}$ $+ \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}}$ $= \frac{α t + β}{t^{2} + 1} + \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}} \Rightarrow$ $\frac{1}{(t^{2} + 1) (t^{n} + 1)} - \frac{α t + β}{t^{2} + 1} = \sum_{p = 0}^{n - 1} \frac{a_{p}}{t - z_{p}} \Rightarrow$ $\frac{1}{(t^{2} + 1)} {\frac{1}{t^{n} + 1} - (α t + β)} = Σ (. . .) \Rightarrow$ $\frac{1 - (α t + β) (t^{n} + 1)}{(t^{2} + 1) (t^{n} + 1)} = \sum_{p = 0}^{\infty} \frac{a_{p}}{t - z_{p}}$ $a_{p} = \frac{f (z_{p)}}{g^{^{'}} (z_{p})} we have f (z_{p}) = \frac{1 - (α z_{p} + β) (z_{p}^{n} + 1)}{(z_{p}^{2} + 1) (z_{p}^{n} + 1)}$ $g (t) = t^{n + 2} + t^{2} + t^{n} + 1 \Rightarrow$ $g^{^{'}} (t) = (n + 2) t^{n + 1} + {nt}^{n - 1} + 2 t \Rightarrow$ $g^{^{'}} (z_{p}) = (n + 2) z_{p}^{n + 1} + {nz}_{p}^{n - 1} + {2 z}_{p} \Rightarrow$ $\int F (t) dt = aln (t - i) + bln (t + i) + \sum_{p = 0}^{n - 1} a_{p} \ln (t - z_{p}) + C$
	Commented by 175 last updated on 04/Jun/20
	nice
	Commented by mathmax by abdo last updated on 04/Jun/20

	$thankx$
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