A = Σ_(n=2) ^(2017) [∫_1 ^n 2tan^(−1) x + sin^(−1) (((2x)/(1 + x^2 ))) dx] B = Π_(n=2) ^(2017) [∫_1 ^n 2tan^(−1) x + sin^(−1) (((2x)/(1 + x^2 ))) dx] A + B = ...


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Question Number 33570 by Joel578 last updated on 19/Apr/18

$A = \underset{n = 2}{\sum^{2017}} [\underset{1}{\int^{n}} {2 \tan}^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x]$ $B = \underset{n = 2}{\prod^{2017}} [\underset{1}{\int^{n}} {2 \tan}^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x]$ $A + B = . . .$
Commented by MJS last updated on 19/Apr/18

$A = B, yes ?$
Commented by Joel578 last updated on 20/Apr/18

$A and B have different symbol Σ and Π$ $But I dont know if they lead to the same result$ $I will try it later$
Commented by MJS last updated on 20/Apr/18

$sorry, I {didn}^{'} t notice$
	Answered by MJS last updated on 20/Apr/18
	$θ=2tan^(−1) x ⇒ x=tan (θ/2) ∧ −π≤θ≤π sin α=((2tan (α/2))/(1+tan^2 (α/2))) ⇒ sin^(−1) ((2x)/(1+x^2 ))=θ ∧ −(π/2)≤θ≤(π/2) ⇒ { ((π+2tan^(−1) x=−sin^(−1) ((2x)/(1+x^2 )); x∈]−∞;−1])),((2tan^(−1) x=sin^(−1) ((2x)/(1+x^2 )); x∈[−1;1])),((π−2tan^(−1) x=sin^(−1) ((2x)/(1+x^2 )); x∈[1;∞[)) :} ⇒ { ((2tan^(−1) x + sin^(−1) ((2x)/(1 + x^2 ))=−π; x∈]−∞;−1])),((2tan^(−1) x + sin^(−1) ((2x)/(1 + x^2 ))4tan^(−1) x; x∈[−1;1])),((2tan^(−1) x + sin^(−1) ((2x)/(1 + x^2 ))π; x∈[1;∞[)) :} ⇒ A=Σ_(n=2) ^(2017) ∫_1 ^n πdx=Σ_(n=2) ^(2017) [πx]_1 ^n =Σ_(n=2) ^(2017) π(n−1)=Σ_(n=1) ^(2016) πn=2033136π B=Π_(n=1) ^(2016) πn=2016!π^(2016) ⇒ A+B=2033136π+2016!π^(2016)$
	$θ = {2 \tan}^{- 1} x \Rightarrow x = \tan \frac{θ}{2} \land - π ⩽ θ ⩽ π$ $\sin α = \frac{2 \tan \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} \Rightarrow \sin^{- 1} \frac{2 x}{1 + x^{2}} = θ \land - \frac{π}{2} ⩽ θ ⩽ \frac{π}{2}$ $\Rightarrow {\begin{cases} π + {2 \tan}^{- 1} x = - \sin^{- 1} \frac{2 x}{1 + x^{2}}; x \in] - \infty; - 1] \\ {2 \tan}^{- 1} x = \sin^{- 1} \frac{2 x}{1 + x^{2}}; x \in [- 1; 1] \\ π - {2 \tan}^{- 1} x = \sin^{- 1} \frac{2 x}{1 + x^{2}}; x \in [1; \infty [ \end{cases}$ $\Rightarrow {\begin{cases} 2 \tan^{- 1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}} = - π; x \in] - \infty; - 1] \\ {2 \tan}^{- 1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}} {4 \tan}^{- 1} x; x \in [- 1; 1] \\ {2 \tan}^{- 1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}} π; x \in [1; \infty [ \end{cases}$ $\Rightarrow A = \underset{n = 2}{\sum^{2017}} \underset{1}{\int^{n}} π d x = \underset{n = 2}{\sum^{2017}} {[π x]}_{1}^{n} = \underset{n = 2}{\sum^{2017}} π (n - 1) = \underset{n = 1}{\sum^{2016}} π n = 2033136 π$ $B = \underset{n = 1}{\prod^{2016}} π n = 2016! π^{2016}$ $\Rightarrow A + B = 2033136 π + 2016! π^{2016}$
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