Question Number 100026 by Ar Brandon last updated on 24/Jun/20

∫_0 ^(π/2) e^(−sec^2 θ) dθ

Answered by mathmax by abdo last updated on 24/Jun/20

I =∫_0 ^(π/(2 ))  e^(−(1/(cos^2 θ)))   dθ    we have 1+tan^2 θ =(1/(cos^2 θ)) ⇒ I =∫_0 ^(π/2)  e^(−(1+tan^2 θ))  dθ  changement tanθ =x  give I =∫_0 ^∞   e^(−(1+x^2 ))  (dx/(1+x^2 ))  = ∫_0 ^∞   (e^(−(1+x^2 )) /(1+x^2 ))dx  let f(a) =∫_0 ^∞   (e^(−a(1+x^2 )) /(1+x^2 ))dx     (a>0) ⇒  f^′ (a) =−∫_0 ^∞  e^(−a(1+x^2 )) dx =−e^(−a)  ∫_0 ^∞  e^(−ax^2 ) dx =_(x(√a)=u)   −e^(−a)  ∫_0 ^∞  e^(−u^2 ) (du/(√a))  =−(e^(−a) /(√a)) ∫_0 ^∞  e^(−u^2 )  du =−((√π)/2) ×(e^(−a) /(√a)) ⇒f(a) =−((√π)/2) ∫_0 ^a  (e^(−t) /(√t))dt +c  =_((√t)=α)  −((√π)/2) ∫_0 ^(√a)  (e^(−α^2 ) /α)(2α)dα +c =−(√π)∫_0 ^(√a)  e^(−α^2 ) dα +c  c=lim_(a→0)   f(a) =(π/2) ⇒f(a) =(π/2)−(√π)∫_0 ^(√a) e^(−α^2 )  dα ⇒  I =f(1) =(π/2)−(√π)∫_0 ^1  e^(−α^2 ) dα  but e^(−α^2 )  =Σ_(n=0) ^∞  (((−α^2 )^n )/(n!)) =Σ_(n=0) ^∞  (((−1)^n )/(n!)) α^(2n)  ⇒  I =(π/2)−(√π)Σ_(n=0) ^∞  (((−1)^n )/(n!)) ∫_0 ^1  α^(2n)  dα  I=(π/2)−(√π)Σ_(n=0) ^∞  (((−1)^n )/((2n+1)(n!)))

Commented byAr Brandon last updated on 24/Jun/20

wow ! thank you

Commented bymathmax by abdo last updated on 24/Jun/20

you are welcome