Question Number 100967 by mathmax by abdo last updated on 29/Jun/20

calculate ∫_0 ^(π/2) ln(2+ sinθ)dθ

Answered by mathmax by abdo last updated on 01/Jul/20

I =∫_0 ^(π/2) ln(2+sinθ)dθ ⇒I =∫_0 ^(π/2) (ln(2)+ln(1+(1/2)sinθ))dθ =(π/2)ln(2) +∫_0 ^(π/2) ln(1+(1/2)sinθ)dθ let f(a) =∫_0 ^(π/2) ln(1+asinθ)dθ with 0<a<1 we have f^′ (a) =∫_0 ^(π/2) ((sinθ)/(1+asinθ)) =(1/a)∫_0 ^(π/2) ((1+asinθ −1)/(1+asinθ))dθ =(π/(2a)) −(1/a) ∫_0 ^(π/2) (dθ/(1+asinθ)) =_(tan((θ/2))=t) (π/(2a))−(1/a) ∫_0 ^1 ((2dt)/((1+t^2 )(1+a×((2t)/(1+t^2 ))))) =(π/(2a)) −(2/a) ∫_0 ^1 (dt/(1+t^2 +2at)) we have ∫_0 ^1 (dt/(t^2 +2at +1)) =∫_0 ^1 (dt/(t^2 +2at +a^2 +1−a^2 )) =∫_0 ^1 (dt/((t+a)^2 +1−a^2 )) =_(t+a =(√(1−a^2 ))u) ∫_(a/(√(1−a^2 ))) ^((1+a)/(√(1−a^2 ))) (((√(1−a^2 ))du)/((1−a^2 )(1+u^2 ))) =(((√(1−a))×(√(1+a)))/((1−a)(1+a))) ∫_(a/(√(1−a^2 ))) ^((1+a)/(√(1−a^2 ))) (du/(1+u^2 )) =(1/(√(1−a^2 ))) { arctan(((1+a)/(√(1−a^2 ))))−arctn((a/(√(1−a^2 ))))} ⇒ f(a) =∫_0 ^a (1/(√(1−z^2 ))) arctan(((1+z)/(√(1−z^2 ))))dz−∫_0 ^a (1/(√(1−z^2 ))) arctan((z/(√(1−z^2 )))) +c c =f(0) =0 ⇒ f(a) =∫_0 ^a (1/(√(1−z^2 ))) arctan(((1+z)/(√(1−z^2 ))))dz −∫_0 ^a (1/(√(1−z^2 ))) arctan((z/(√(1−z^2 ))))dz I =(π/2)ln(2)+f((1/2)) =(π/2)ln(2) +∫_0 ^(1/2) (1/(√(1−z^2 ))) arctan(((1+z)/(√(1−z^2 ))))−∫_0 ^(1/(2 )) (1/(√(1−z^2 ))) arctan((z/(√(1−z^2 ))))dz rest calculus of this integrals....