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Question Number 36938 by maxmathsup by imad last updated on 07/Jun/18
1) find f(a) = ∫_0 ^(π/2) (dt/(1+a cost)) 2) find A(θ) =∫_0 ^(π/2) (dt/(1+sinθ cost))
1)findf(a)=∫0π2dt1+acost2)findA(θ)=∫0π2dt1+sinθcost
Commented by math khazana by abdo last updated on 09/Jun/18
1) changemnt tan((t/2)) =x give f(a) = ∫_0 ^1 (1/(1+a ((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) = 2 ∫_0 ^1 (dx/(1+x^2 +a −ax^2 )) = 2 ∫_0 ^1 (dx/((1−a)x^2 +1+a)) = (2/(1−a))∫_0 ^1 (dx/(x^2 +((1+a)/(1−a)))) case1 ((1+a)/(1−a))> o ⇒∣a∣<1 we get by chang.x=(√((1+a)/(1−a))) u f(a) = (2/(1−a)) ∫_0 ^(√((1−a)/(1+a))) (1/(((1+a)/(1−a))(1+u^2 ))) (√((1+a)/(1−a))) du = (2/(1−a)) ((√(1−a))/( (√(1+a)))) [arctan(u)]_0 ^(√((1−a)/(1+a))) = (2/( (√(1−a^2 )))) arctan((√((1−a)/(1+a)))) case 2 if ((1+a)/(1−a))<0 ⇒ ∣a∣>1 ∫_0 ^1 (dx/(x^2 +((1+a)/(1−a)))) =∫_0 ^1 (dx/(x^2 −((a+1)/(a−1)))) =_(x =(√((a+1)/(a−1))) u) ∫_0 ^(√((a−1)/(a+1))) (1/(((a+1)/(a−1))(u^2 −1))) (√((a+1)/(a−1))) du =(1/2)((√(a−1))/( (√(a+1)))) ∫_0 ^(√((a−1)/(a+1))) { (1/(u−1)) −(1/(u+1))}du = ((√(a−1))/(2(√(a+1)))) [ln∣((u−1)/(u+1))∣]_0 ^(√((a−1)/(a+1))) =((√(a−1))/(2(√(a+1)))) ln∣ (((√((a−1)/(a+1))) −1)/( (√((a−1)/(a+1))) +1))∣ ⇒ f(a) = (2/(1−a)) ((√(a−1))/(2(√(a+1)))) ln∣ (((√((a−1)/(a+1))) −1)/( (√((a−1)/(a+1))) +1))∣ = ((−1)/( (√(a^2 −1))))ln∣ (((√((a−1)/(a+1))) −1)/( (√((a−1)/(a+1))) +1)) ∣
1)changemnttan(t2)=xgivef(a)=∫0111+a1−x21+x22dx1+x2=2∫01dx1+x2+a−ax2=2∫01dx(1−a)x2+1+a=21−a∫01dxx2+1+a1−acase11+a1−a>o⇒∣a∣<1wegetbychang.x=1+a1−auf(a)=21−a∫01−a1+a11+a1−a(1+u2)1+a1−adu=21−a1−a1+a[arctan(u)]01−a1+a=21−a2arctan(1−a1+a)case2if1+a1−a<0⇒∣a∣>1∫01dxx2+1+a1−a=∫01dxx2−a+1a−1=x=a+1a−1u∫0a−1a+11a+1a−1(u2−1)a+1a−1du=12a−1a+1∫0a−1a+1{1u−1−1u+1}du=a−12a+1[ln∣u−1u+1∣]0a−1a+1=a−12a+1ln∣a−1a+1−1a−1a+1+1∣⇒f(a)=21−aa−12a+1ln∣a−1a+1−1a−1a+1+1∣=−1a2−1ln∣a−1a+1−1a−1a+1+1∣
Commented by math khazana by abdo last updated on 09/Jun/18
2) we have ∣sinθ∣≤1 so if ∣sinθ∣ <1 ∫_0 ^(π/2) (dt/(1+cosθ cost))=f(sinθ) =(2/( (√(1−sin^2 θ)))) arctan{ (√((1−sinθ)/(1+sinθ)))} = (2/(∣cosθ∣)) arctan(√((1−sinθ)/(1+sinθ)))
2)wehave∣sinθ∣⩽1soif∣sinθ∣<1∫0π2dt1+cosθcost=f(sinθ)=21−sin2θarctan{1−sinθ1+sinθ}=2∣cosθ∣arctan1−sinθ1+sinθ

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