0-pi-2-arc-tan-tan-x-tan-x-dx-

Question Number 86167 by jagoll last updated on 27/Mar/20

∫_0 ^(π/2) ((arc tan ((√(tan x))))/(tan x)) dx

\underset{0}{\int^{\frac{π}{2}}} \frac{arc \tan (\sqrt{\tan x})}{\tan x} dx

Commented by mathmax by abdo last updated on 27/Mar/20

let take a try I =∫_0 ^(π/2) ((arctan((√(tanx))))/(tanx))dx changement (√(tanx))=t give tanx=t^2 ⇒ x =arctan(t^2 ) ⇒ I =∫_0 ^(+∞) ((arctan(t))/t^2 )×((2t)/(1+t^4 ))dt =2 ∫_0 ^∞ ((arctant)/(t(1+t^4 )))dt the convergence of this integral is assured let f(a) =∫_0 ^∞ ((arctan(at))/(t(1+t^4 )))dt with a>0 f^′ (a) =∫_0 ^∞ (t/(t(1+a^2 t^2 )(1+t^4 )))dt =∫_0 ^∞ (dt/((t^4 +1)(1+a^2 t^2 ))) =_(at=u) ∫_0 ^∞ (du/(a((u^4 /a^4 ) +1)(1+u^2 ))) =∫_0 ^∞ ((a^3 du)/((u^2 +1)(u^4 +a^4 ))) ⇒ 2f^′ (a) =a^3 ∫_(−∞) ^(+∞) (du/((u^2 +1)(u^4 +a^4 ))) let ϕ(z) =(1/((z^2 +1)(z^4 +a^4 ))) ⇒ ϕ(z) =(1/((z−i)(z+i)(z^2 −ia^2 )(z^2 +ia^2 ))) =(1/((z−i)(z+i)(z−ae^((iπ)/4) )(z+ae^((iπ)/4) )(z−ae^(−((iπ)/4)) )(z+a e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,i) +Res(ϕ,ae^((iπ)/4) ) +Re(ϕ,−ae^(−((iπ)/4)) )} Res(ϕ,i) =(1/((2i)(1+a^4 ))) Res(ϕ,ae^((iπ)/4) ) =(1/((a^2 i+1)(2ae^((iπ)/4) )(2ia^2 ))) =(e^(−((iπ)/4)) /(4ia^3 (1+a^2 i))) Res(ϕ,−ae^(−((iπ)/4)) ) =(1/((−a^2 i+1)(−2ia^2 )(−2ae^(−((iπ)/4)) ))) =(e^((iπ)/4) /(4ia^3 (1−a^2 i))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/((2i)(1+a^4 ))) +(e^(−((iπ)/4)) /(4ia^3 (1+a^2 i))) +(e^((iπ)/4) /(4ia^3 (1−a^2 i)))} =(π/(1+a^4 )) +(1/(2a^3 ))( (e^(−((iπ)/4)) /(1+a^2 i)) +(e^((iπ)/4) /(1−a^2 i))) =(π/(1+a^4 )) +(1/(2a^3 ))(2Re((e^((iπ)/4) /(1−a^2 i)))) =(π/(1+a^4 )) +(1/a^3 )Re((e^((iπ)/4) /(1−a^2 i))) (e^((iπ)/4) /(1−a^2 i)) =(((1+a^2 i) e^((iπ)/4) )/(1+a^4 )) =(((1+a^2 i)((1/( (√2)))+(i/( (√2)))))/(1+a^4 )) =(1/( (√2)(1+a^4 )))(1+a^2 i)(1+i) =(1/( (√2)(1+a^4 )))(1+i+a^2 i−a^2 ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =(π/(1+a^4 )) +((1−a^2 )/(a^3 (√2)(1+a^4 ))) ⇒ 2f^′ (a) =((πa^3 )/(1+a^4 )) +((1−a^2 )/( (√2)(1+a^4 ))) ⇒f^′ (a) =(π/2)×(a^3 /(1+a^4 )) +(1/(2(√2)))×((1−a^2 )/(1+a^4 )) ⇒ f(a) =(π/2)∫_0 ^a (x^3 /(1+x^4 ))dx +(1/(2(√2))) ∫_0 ^a ((1−x^2 )/(1+x^4 ))dx +C C=f(0) =0 ⇒f(a) =(π/8)ln(1+a^4 )+(1/(2(√2)))∫_0 ^a ((1−x^2 )/(1+x^4 ))dx I =2f(1) ....be continued....

l e t t a k e a t r y

I = \int_{0}^{\frac{π}{2}} \frac{a r c t a n (\sqrt{t a n x})}{t a n x} d x c h a n g e m e n t \sqrt{t a n x} = t g i v e t a n x = t^{2} \Rightarrow

x = a r c t a n (t^{2}) \Rightarrow I = \int_{0}^{+ \infty} \frac{a r c t a n (t)}{t^{2}} \times \frac{2 t}{1 + t^{4}} d t

= 2 \int_{0}^{\infty} \frac{a r c t a n t}{t (1 + t^{4})} d t t h e c o n v e r g e n c e o f t h i s i n t e g r a l i s a s s u r e d

l e t f (a) = \int_{0}^{\infty} \frac{a r c t a n (a t)}{t (1 + t^{4})} d t w i t h a > 0

f^{^{'}} (a) = \int_{0}^{\infty} \frac{t}{t (1 + a^{2} t^{2}) (1 + t^{4})} d t = \int_{0}^{\infty} \frac{d t}{(t^{4} + 1) (1 + a^{2} t^{2})}

=_{a t = u} \int_{0}^{\infty} \frac{d u}{a (\frac{u^{4}}{a^{4}} + 1) (1 + u^{2})} = \int_{0}^{\infty} \frac{a^{3} d u}{(u^{2} + 1) (u^{4} + a^{4})} \Rightarrow

2 f^{^{'}} (a) = a^{3} \int_{- \infty}^{+ \infty} \frac{d u}{(u^{2} + 1) (u^{4} + a^{4})} l e t φ (z) = \frac{1}{(z^{2} + 1) (z^{4} + a^{4})} \Rightarrow

φ (z) = \frac{1}{(z - i) (z + i) (z^{2} - {i a}^{2}) (z^{2} + {i a}^{2})}

= \frac{1}{(z - i) (z + i) (z - {a e}^{\frac{i π}{4}}) (z + {a e}^{\frac{i π}{4}}) (z - {a e}^{- \frac{i π}{4}}) (z + a e^{- \frac{i π}{4}})}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, {a e}^{\frac{i π}{4}}) + R e (φ, - {a e}^{- \frac{i π}{4}})}

R e s (φ, i) = \frac{1}{(2 i) (1 + a^{4})}

R e s (φ, {a e}^{\frac{i π}{4}}) = \frac{1}{(a^{2} i + 1) (2 {a e}^{\frac{i π}{4}}) (2 {i a}^{2})} = \frac{e^{- \frac{i π}{4}}}{4 {i a}^{3} (1 + a^{2} i)}

R e s (φ, - {a e}^{- \frac{i π}{4}}) = \frac{1}{(- a^{2} i + 1) (- 2 {i a}^{2}) (- 2 {a e}^{- \frac{i π}{4}})} = \frac{e^{\frac{i π}{4}}}{4 {i a}^{3} (1 - a^{2} i)} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{(2 i) (1 + a^{4})} + \frac{e^{- \frac{i π}{4}}}{4 {i a}^{3} (1 + a^{2} i)} + \frac{e^{\frac{i π}{4}}}{4 {i a}^{3} (1 - a^{2} i)}}

= \frac{π}{1 + a^{4}} + \frac{1}{2 a^{3}} (\frac{e^{- \frac{i π}{4}}}{1 + a^{2} i} + \frac{e^{\frac{i π}{4}}}{1 - a^{2} i})

= \frac{π}{1 + a^{4}} + \frac{1}{2 a^{3}} (2 R e (\frac{e^{\frac{i π}{4}}}{1 - a^{2} i})) = \frac{π}{1 + a^{4}} + \frac{1}{a^{3}} R e (\frac{e^{\frac{i π}{4}}}{1 - a^{2} i})

\frac{e^{\frac{i π}{4}}}{1 - a^{2} i} = \frac{(1 + a^{2} i) e^{\frac{i π}{4}}}{1 + a^{4}} = \frac{(1 + a^{2} i) (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}{1 + a^{4}}

= \frac{1}{\sqrt{2} (1 + a^{4})} (1 + a^{2} i) (1 + i) = \frac{1}{\sqrt{2} (1 + a^{4})} (1 + i + a^{2} i - a^{2}) \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{1 + a^{4}} + \frac{1 - a^{2}}{a^{3} \sqrt{2} (1 + a^{4})} \Rightarrow

2 f^{^{'}} (a) = \frac{π a^{3}}{1 + a^{4}} + \frac{1 - a^{2}}{\sqrt{2} (1 + a^{4})} \Rightarrow f^{^{'}} (a) = \frac{π}{2} \times \frac{a^{3}}{1 + a^{4}} + \frac{1}{2 \sqrt{2}} \times \frac{1 - a^{2}}{1 + a^{4}} \Rightarrow

f (a) = \frac{π}{2} \int_{0}^{a} \frac{x^{3}}{1 + x^{4}} d x + \frac{1}{2 \sqrt{2}} \int_{0}^{a} \frac{1 - x^{2}}{1 + x^{4}} d x + C

C = f (0) = 0 \Rightarrow f (a) = \frac{π}{8} l n (1 + a^{4}) + \frac{1}{2 \sqrt{2}} \int_{0}^{a} \frac{1 - x^{2}}{1 + x^{4}} d x

I = 2 f (1) \dots . b e c o n t i n u e d \dots .

Commented by mathmax by abdo last updated on 27/Mar/20

let try anther way we have I =2∫_0 ^∞ ((arctant)/(t(1+t^4 )))dt ⇒I =∫_(−∞) ^(+∞) ((arctant)/(t(t^4 +1)))dt let ϕ(z) =((arctanz)/(z(z^4 +1))) ⇒ ϕ(z) =((arctanz)/(z(z^2 −i)(z^2 +i))) =((arctan(z))/(z(z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,0) +Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,0)=0 Res(ϕ,e^((iπ)/4) ) =((arctan(e^((iπ)/4) ))/(e^((iπ)/4) (2e^((iπ)/4) )(2i))) =((arctan(e^((iπ)/4) ))/(4i (i))) =−(1/4) arctan(e^((iπ)/4) ) Res(ϕ,−e^(−((iπ)/4)) ) =((−arctan(e^(−((iπ)/4)) ))/((−e^(−((iπ)/4)) )(−2e^(−((iπ)/4)) )(−2i))) =((arctan(e^(−((iπ)/4)) ))/(4i(−i))) =(1/4) arctan(e^(−((iπ)/4)) ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{−(1/4)arctan(e^((iπ)/4) )+(1/4) arctan(e^(−((iπ)/4)) )} =−((iπ)/2){ arctan(e^((iπ)/4) )−arctan(e^(−((iπ)/4)) )} we know arctan(z) =(1/(2i))ln(((1+iz)/(1−iz))) and arctan(z^− )=(1/(2i))ln(((1+iz^− )/(1−iz^− ))) ⇒arctan(z)−arctan(z^− ) =(1/(2i))ln(((1+iz)/(1−iz))×((1−iz^− )/(1+iz^− ))) =(1/(2i))ln(((1−iz^− +iz +zz^− )/(1+iz^− −iz +zz^− ))) =(1/(2i))ln(((1+i(2iIm(z)+∣z∣^2 )/(1−i(2iIm(z))+∣z∣^2 ))) =(1/(2i))ln(((1−2Im(z)+∣z∣^2 )/(1+2Im(z)+∣z∣^2 ))) ⇒ arctan(e^((iπ)/4) )−arctan(e^(−((iπ)/4)) ) =(1/(2i))ln(((1−2×((√2)/2)+1)/(1+2×((√2)/2)+1))) =(1/(2i))ln(((2−(√2))/(2+(√2)))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =−((iπ)/2)×(1/(2i))ln(((2−(√2))/(2+(√2)))) =−(π/4)ln(((2−(√2))/(2+(√2)))) =(π/4)ln(((2+(√2))/(2−(√2)))) =I

l e t t r y a n t h e r w a y w e h a v e I = 2 \int_{0}^{\infty} \frac{a r c t a n t}{t (1 + t^{4})} d t

\Rightarrow I = \int_{- \infty}^{+ \infty} \frac{a r c t a n t}{t (t^{4} + 1)} d t l e t φ (z) = \frac{a r c t a n z}{z (z^{4} + 1)} \Rightarrow

φ (z) = \frac{a r c t a n z}{z (z^{2} - i) (z^{2} + i)} = \frac{a r c t a n (z)}{z (z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, 0) + R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}

R e s (φ, 0) = 0

R e s (φ, e^{\frac{i π}{4}}) = \frac{a r c t a n (e^{\frac{i π}{4}})}{e^{\frac{i π}{4}} (2 e^{\frac{i π}{4}}) (2 i)} = \frac{a r c t a n (e^{\frac{i π}{4}})}{4 i (i)} = - \frac{1}{4} a r c t a n (e^{\frac{i π}{4}})

R e s (φ, - e^{- \frac{i π}{4}}) = \frac{- a r c t a n (e^{- \frac{i π}{4}})}{(- e^{- \frac{i π}{4}}) (- 2 e^{- \frac{i π}{4}}) (- 2 i)} = \frac{a r c t a n (e^{- \frac{i π}{4}})}{4 i (- i)}

= \frac{1}{4} a r c t a n (e^{- \frac{i π}{4}}) \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {- \frac{1}{4} a r c t a n (e^{\frac{i π}{4}}) + \frac{1}{4} a r c t a n (e^{- \frac{i π}{4}})}

= - \frac{i π}{2} {a r c t a n (e^{\frac{i π}{4}}) - a r c t a n (e^{- \frac{i π}{4}})}

w e k n o w a r c t a n (z) = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z}) a n d a r c t a n (\bar{z}) = \frac{1}{2 i} l n (\frac{1 + i \bar{z}}{1 - i \bar{z}})

\Rightarrow a r c t a n (z) - a r c t a n (\bar{z}) = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z} \times \frac{1 - i \bar{z}}{1 + i \bar{z}})

= \frac{1}{2 i} l n (\frac{1 - i \bar{z} + i z + z \bar{z}}{1 + i \bar{z} - i z + z \bar{z}}) = \frac{1}{2 i} l n (\frac{1 + i (2 i I m (z) + ∣ z ∣^{2}}{1 - i (2 i I m (z)) + ∣ z ∣^{2}})

= \frac{1}{2 i} l n (\frac{1 - 2 I m (z) + ∣ z ∣^{2}}{1 + 2 I m (z) + ∣ z ∣^{2}}) \Rightarrow

a r c t a n (e^{\frac{i π}{4}}) - a r c t a n (e^{- \frac{i π}{4}})

= \frac{1}{2 i} l n (\frac{1 - 2 \times \frac{\sqrt{2}}{2} + 1}{1 + 2 \times \frac{\sqrt{2}}{2} + 1}) = \frac{1}{2 i} l n (\frac{2 - \sqrt{2}}{2 + \sqrt{2}}) \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = - \frac{i π}{2} \times \frac{1}{2 i} l n (\frac{2 - \sqrt{2}}{2 + \sqrt{2}}) = - \frac{π}{4} l n (\frac{2 - \sqrt{2}}{2 + \sqrt{2}})

= \frac{π}{4} l n (\frac{2 + \sqrt{2}}{2 - \sqrt{2}}) = I

Commented by M±th+et£s last updated on 28/Mar/20

∫_0 ^(π/2) ((tan^(−1) ((√(tan(x)))))/(tan(x))) dx y^2 =tan(x) tan^− (y^2 )=x dx=((2y)/(1+y^4 )) dy I=∫_0 ^∞ ((tan^(−1) (y))/y^2 ) . ((2y)/(1+y^4 )) dy=∫_0 ^∞ ((2 tan^(−1) (y))/(y(1+y^4 )))dy I(a)=∫_0 ^∞ ((2tan^(−1) (ay))/(y(1+y^4 )))dy I(a)=∫_0 ^∞ (((2y)/(1+a^2 y^2 ))/(y(1+y^4 )))dy=∫_0 ^∞ (2/((1+a^2 y^2 )(1+y^4 )))dy =(2/(a^4 +1))∫_0 ^∞ ((a^2 /(1+a^2 y^2 ))+((1−a^2 y^2 )/(1+y^4 )))dy =(2/(a^4 +1))[∫_0 ^∞ (a^2 /(1+a^2 y^2 ))dy + ∫_0 ^∞ (1/(1+y^4 ))dy −a^2 ∫_0 ^∞ (y^2 /(1+y^4 ))dy] (2/(a^4 +1))[((a^3 π)/2)+(π/(2(√2)))−((a^2 π)/(2(√2)))] ((2π)/(a^4 +1))[((a^4 −(√2) a^2 +(√2))/4)]=(π/2)[((2a^3 −(√2)a^2 +(√2))/((a^2 +1)^2 −2a^2 ))] =(π/2)[(((2a+(√2))(a^2 −(√2)a+1))/((a^2 +(√2)a+1)(a^2 −(√2)a+1)))] I(0)=∫_0 ^∞ ((2tan^(−1) (0))/(y(1+y^4 )))dy=0 ⇒⇒0=(π/2)ln(0+0+1)+c c=0 I(a)=(π/2)ln(a^2 +(√2) a+1) I(1)=(π/2)ln(2+(√2)) notice// ∫_0 ^∞ (dy/(1+y^4 ))=(π/(2(√2))) and ∫_0 ^∞ (y^2 /(1+y^4 ))dy=(π/(2(√2))) and ∫_0 ^∞ (dy/(1+a^2 y^2 ))=((aπ)/2)

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

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

I = \int_{0}^{\infty} \frac{{t a n}^{- 1} (y)}{y^{2}} . \frac{2 y}{1 + y^{4}} d y = \int_{0}^{\infty} \frac{2 {t a n}^{- 1} (y)}{y (1 + y^{4})} d y

I (a) = \int_{0}^{\infty} \frac{2 {t a n}^{- 1} (a y)}{y (1 + y^{4})} d y

I (a) = \int_{0}^{\infty} \frac{\frac{2 y}{1 + a^{2} y^{2}}}{y (1 + y^{4})} d y = \int_{0}^{\infty} \frac{2}{(1 + a^{2} y^{2}) (1 + y^{4})} d y

= \frac{2}{a^{4} + 1} \int_{0}^{\infty} (\frac{a^{2}}{1 + a^{2} y^{2}} + \frac{1 - a^{2} y^{2}}{1 + y^{4}}) d y

= \frac{2}{a^{4} + 1} [\int_{0}^{\infty} \frac{a^{2}}{1 + a^{2} y^{2}} d y + \int_{0}^{\infty} \frac{1}{1 + y^{4}} d y - a^{2} \int_{0}^{\infty} \frac{y^{2}}{1 + y^{4}} d y]

\frac{2}{a^{4} + 1} [\frac{a^{3} π}{2} + \frac{π}{2 \sqrt{2}} - \frac{a^{2} π}{2 \sqrt{2}}]

\frac{2 π}{a^{4} + 1} [\frac{a^{4} - \sqrt{2} a^{2} + \sqrt{2}}{4}] = \frac{π}{2} [\frac{2 a^{3} - \sqrt{2} a^{2} + \sqrt{2}}{{(a^{2} + 1)}^{2} - 2 a^{2}}]

= \frac{π}{2} [\frac{(2 a + \sqrt{2}) (a^{2} - \sqrt{2} a + 1)}{(a^{2} + \sqrt{2} a + 1) (a^{2} - \sqrt{2} a + 1)}]

I (0) = \int_{0}^{\infty} \frac{2 {t a n}^{- 1} (0)}{y (1 + y^{4})} d y = 0 \Rightarrow\Rightarrow 0 = \frac{π}{2} l n (0 + 0 + 1) + c c = 0

I (a) = \frac{π}{2} l n (a^{2} + \sqrt{2} a + 1)

I (1) = \frac{π}{2} l n (2 + \sqrt{2})

n o t i c e / / \int_{0}^{\infty} \frac{d y}{1 + y^{4}} = \frac{π}{2 \sqrt{2}} a n d \int_{0}^{\infty} \frac{y^{2}}{1 + y^{4}} d y = \frac{π}{2 \sqrt{2}} a n d \int_{0}^{\infty} \frac{d y}{1 + a^{2} y^{2}} = \frac{a π}{2}

Answered by TANMAY PANACEA. last updated on 27/Mar/20

(√(tanx)) =tant tanx=tan^2 t sec^2 xdx=2tant.sec^2 tdt dx=((2tantsec^2 t)/(1+tan^4 t))dt ∫_0 ^(π/2) ((2tantsec^2 t)/(1+tan^4 t))×(t/(tan^2 t))dt (I/2)=∫_0 ^(π/2) t×(1/(cos^2 t×((sint)/(cost))))×((cos^4 t)/(sin^4 t+cos^4 t))dt (I/2)=∫_0 ^(π/2) ((π/2)−t)×(1/(sintcost))×((sin^4 t)/(cos^4 t+sin^4 t))dt using ∫_0 ^a f(x)dx=∫_0 ^a f(a−x)dx (I/2)+(I/2)=∫_0 ^(π/2) (π/2)×((sin^4 t)/(sintcost))×(1/(sin^4 t+cos^4 t)) ((2I)/π)=∫_0 ^(π/2) (1/(tant))×sec^2 t×(1/(1+tan^4 t))×dt a=tant ((2I)/π)=∫_0 ^∞ (da/(a(1+a^4 ))) =(1/2)∫_0 ^∞ ((d(a^2 ))/(a^2 (1+a^4 ))) replace a^2 by b ((4I)/π)=∫_0 ^∞ (db/(b(1+b^2 )))=(1/2)∫_0 ^∞ ((2bdb)/(b^2 (1+b^2 ))) ((8I)/π)=∫_0 ^∞ (dk/(k(1+k))) [k=b^2 ] =∫_0 ^∞ ((k+1−k)/(k(1+k)))dk =∫_0 ^∞ (dk/k)−∫_0 ^∞ (dk/(k+1)) =∣ln((k/(k+1)))∣_0 ^∞ some thing wrong

\sqrt{t a n x} = t a n t

t a n x = {t a n}^{2} t

{s e c}^{2} x d x = 2 t a n t . {s e c}^{2} t d t

d x = \frac{2 {t a n t s e c}^{2} t}{1 + {t a n}^{4} t} d t

\int_{0}^{\frac{π}{2}} \frac{2 {t a n t s e c}^{2} t}{1 + {t a n}^{4} t} \times \frac{t}{{t a n}^{2} t} d t

\frac{I}{2} = \int_{0}^{\frac{π}{2}} t \times \frac{1}{{c o s}^{2} t \times \frac{s i n t}{c o s t}} \times \frac{{c o s}^{4} t}{{s i n}^{4} t + {c o s}^{4} t} d t

\frac{I}{2} = \int_{0}^{\frac{π}{2}} (\frac{π}{2} - t) \times \frac{1}{s i n t c o s t} \times \frac{{s i n}^{4} t}{{c o s}^{4} t + {s i n}^{4} t} d t

u s i n g \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

\frac{I}{2} + \frac{I}{2} = \int_{0}^{\frac{π}{2}} \frac{π}{2} \times \frac{{s i n}^{4} t}{s i n t c o s t} \times \frac{1}{{s i n}^{4} t + {c o s}^{4} t}

\frac{2 I}{π} = \int_{0}^{\frac{π}{2}} \frac{1}{t a n t} \times {s e c}^{2} t \times \frac{1}{1 + {t a n}^{4} t} \times d t

a = t a n t

\frac{2 I}{π} = \int_{0}^{\infty} \frac{d a}{a (1 + a^{4})}

= \frac{1}{2} \int_{0}^{\infty} \frac{d (a^{2})}{a^{2} (1 + a^{4})} r e p l a c e a^{2} b y b

\frac{4 I}{π} = \int_{0}^{\infty} \frac{d b}{b (1 + b^{2})} = \frac{1}{2} \int_{0}^{\infty} \frac{2 b d b}{b^{2} (1 + b^{2})}

\frac{8 I}{π} = \int_{0}^{\infty} \frac{d k}{k (1 + k)} [k = b^{2}]

= \int_{0}^{\infty} \frac{k + 1 - k}{k (1 + k)} d k

= \int_{0}^{\infty} \frac{d k}{k} - \int_{0}^{\infty} \frac{d k}{k + 1}

=∣ l n (\frac{k}{k + 1}) ∣_{0}^{\infty}

s o m e t h i n g w r o n g

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