...nice calculus... prove that : φ = ∫_0 ^( π) ((tan^(−1) (tan^2 (x)))/(tan^2 (x)) dx=^(???) π((√2) −(π/4))


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Question Number 128225 by mnjuly1970 last updated on 05/Jan/21

$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t :$ $ϕ = \int_{0}^{π} \frac{{t a n}^{- 1} ({t a n}^{2} (x))}{{t a n}^{2} (x} d x \overset{? ? ?}{=} π (\sqrt{2} - \frac{π}{4})$
	Answered by mathmax by abdo last updated on 05/Jan/21
	$Φ=∫_0 ^π ((arctan(tan^2 x))/(tan^2 x))dx let f(a) =∫_0 ^π ((arctan(atan^2 x))/(tan^2 x))dx (a>0) ⇒ f^′ (a)=∫_0 ^π (1/((1+a^2 tan^4 x)))dx =∫_0 ^(π/2) (...)dx +∫_(π/2) ^π (...)dx ∫_0 ^(π/2) (dx/(1+a^2 tan^4 x)) =_(tanx=t) ∫_0 ^∞ (dt/((1+t^2 )(1+a^2 t^4 ))) =(1/2)∫_(−∞) ^(+∞) (dt/((t^2 +1)(a^2 t^4 +1))) let ϕ(z)=(1/((z^2 +1)(a^2 z^4 +1))) ⇒ϕ(z)=(1/(a^2 (z^2 +1)(z^4 +(1/a^2 )))) =(1/(a^2 (z−i)(z+i)(z^2 −(i/a))(z^2 +(i/a)))) =(1/(a^2 (z−i)(z+i)(z−(1/( (√a)))e^((iπ)/4) )(z+(1/( (√a)))e^((iπ)/4) )(z−(1/( (√a)))e^(−((iπ)/4)) )(z+(1/( (√a)))e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,i) +Res(ϕ,(1/( (√a)))e^((iπ)/4) ) +Res(ϕ,−(1/( (√a)))e^(−((iπ)/4)) )} ...be continued...$
	$Φ = \int_{0}^{π} \frac{\arctan (\tan^{2} x)}{\tan^{2} x} dx let f (a) = \int_{0}^{π} \frac{\arctan ({atan}^{2} x)}{\tan^{2} x} dx (a > 0) \Rightarrow$ $f^{^{'}} (a) = \int_{0}^{π} \frac{1}{(1 + a^{2} \tan^{4} x)} dx = \int_{0}^{\frac{π}{2}} (. . .) dx + \int_{\frac{π}{2}}^{π} (. . .) dx$ $\int_{0}^{\frac{π}{2}} \frac{dx}{1 + a^{2} \tan^{4} x} =_{tanx = t} \int_{0}^{\infty} \frac{dt}{(1 + t^{2}) (1 + a^{2} t^{4})}$ $= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dt}{(t^{2} + 1) (a^{2} t^{4} + 1)} let φ (z) = \frac{1}{(z^{2} + 1) (a^{2} z^{4} + 1)}$ $\Rightarrow φ (z) = \frac{1}{a^{2} (z^{2} + 1) (z^{4} + \frac{1}{a^{2}})} = \frac{1}{a^{2} (z - i) (z + i) (z^{2} - \frac{i}{a}) (z^{2} + \frac{i}{a})}$ $= \frac{1}{a^{2} (z - i) (z + i) (z - \frac{1}{\sqrt{a}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{a}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{a}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{a}} e^{- \frac{i π}{4}})}$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, i) + Res (φ, \frac{1}{\sqrt{a}} e^{\frac{i π}{4}}) + Res (φ, - \frac{1}{\sqrt{a}} e^{- \frac{i π}{4}})}$ $. . . be continued . . .$
	Answered by mindispower last updated on 06/Jan/21

	$t a n (x) = t$ $= \int_{- \infty}^{\infty} \frac{{t a n}^{-} (t^{2})}{t^{2} (1 + t^{2})} d t$ $\int_{- \infty}^{\infty} \frac{{t a n}^{-} (t^{2})}{t^{2}} - \int_{- \infty}^{\infty} \frac{{t a n}^{-} (t^{2})}{1 + t^{2}} d t$ $= 2 (\int_{0}^{\infty} \frac{{t a n}^{-} (t^{2})}{t^{2}} - \int_{0}^{\infty} \frac{{t a n}^{-} (t^{2})}{1 + t^{2}}) d t$ $= 2 A - 2 B$ $A = {[- \frac{{t a n}^{-} (t^{2})}{t}]}_{0}^{\infty} + 2 \int_{0}^{\infty} \frac{d t}{1 + t^{4}} ∣ t^{4} = u$ $= \frac{1}{2} \int_{0}^{\infty} \frac{u^{- \frac{3}{4}}}{1 + u} = \frac{1}{2} β (\frac{1}{4}, \frac{3}{4}) = \frac{π}{2 s i n (\frac{π}{4})} = \frac{π}{\sqrt{2}}$ $B = [{t a n}^{-} (t) {t a n}^{-} (t^{2})] - \int_{0}^{\infty} \frac{2 {t t a n}^{-} (t)}{1 + t^{4}}$ $= \frac{π^{2}}{4} - \int_{0}^{\infty} t^{4} \frac{\frac{2}{t} (\frac{π}{2} - {t a n}^{-} (t))}{1 + t^{4}} . \frac{d t}{t^{2}}_{= \frac{π^{2}}{4} - C}$ $= \frac{π^{2}}{4} - \int_{0}^{\infty} \frac{t π}{1 + t^{4}} + \int_{0}^{\infty} \frac{2 {t t a n}^{-} (t)}{1 + t^{4}} d t$ ${\Rightarrow 2 C = \int_{0}^{\infty} \frac{π t}{1 + t^{4}} = \frac{π}{2} \int_{0}^{\infty} . \frac{d (t^{2})}{1 + {(t^{2})}^{2}} = \frac{π}{2 .} . {t a n}^{-} (t^{2})]}_{0}^{\infty}$ $= \frac{π^{2}}{4} \Rightarrow C = \frac{π^{2}}{8}$ $B = \frac{π^{2}}{8}$ $2 A - 2 B = π \sqrt{2} - \frac{π^{2}}{4} = π (\sqrt{2} - \frac{π}{4}) = \int_{0}^{\frac{π}{2}} \frac{{t a n}^{-} ({t g}^{2} (t))}{{t g}^{2} (t)} d t$
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