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