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....