Question Number 104312 by Ar Brandon last updated on 20/Jul/20

∫_0 ^(π/4) ((√(sin^2 θ+2))/(sinθ))dθ

Answered by mathmax by abdo last updated on 21/Jul/20

I =∫_0 ^(π/4)  ((√(sin^2 θ+2))/(sinθ))dθ ⇒I =∫_0 ^(π/4)  ((sinθ(√(1+(2/(sin^2 θ)))))/(sinθ))dθ =∫_0 ^(π/4) (√(1+(2/(1−cos^2 θ))))dθ  we have 1+tan^2 θ =(1/(cos^2 θ)) ⇒cos^2 θ =(1/(1+tan^2 θ)) ⇒1−cos^2 θ=1−(1/(1+tan^2 θ))  =((tan^2 θ)/(1+tan^2 θ)) ⇒(2/(1−cos^2 θ)) =2×((1+tan^2 θ)/(tan^2 θ)) ⇒  I =∫_0 ^(π/4) (√(1+((2+2tan^2 θ)/(tan^2 θ))))dθ  =∫_0 ^(π/4) (√((3tan^2 θ+2)/(tan^2 θ)))dθ  =_(tanθ =x)   =∫_0 ^1 (√((3x^2 +2)/x^2 )) (dx/(1+x^2 ))  =∫_0 ^1   ((√(3x^2 +2))/(x(1+x^2 )))dx  =(√3)∫_0 ^1  ((√(x^2  +(2/3)))/(x(1+x^2 )))dx  cha7gement x =(√(2/3))shu give u =argsh((√(3/2))x)  I =(√3)∫_0 ^(arhsh((√(3/2))))   (2/3)×((chu)/((√(2/3))sh(u)(1+(2/3)sh^2 u)))×(√(2/3))ch(u)du  =2(√3)∫_0 ^(ln((√(3/2))+(√(1+(9/4)))))    ((ch^2 u)/(shu(3+2sh^2 u)))du    and this integral  can be solved  ...be continued...

Commented byAr Brandon last updated on 21/Jul/20

wow wow wow ! superb. Thanks so much 😃

Commented byAr Brandon last updated on 21/Jul/20

    ((ch^2 u)/(shu(3+2sh^2 u)))=((1+sh^2 u)/(shu(3+2sh^2 u)))  ((1+t^2 )/(t(3+2t^2 )))=((at+b)/(2t^2 +3))+(c/t)=(((at+b)t+c(2t^2 +3))/(t(2t^2 +3)))  t→0 , c=(1/3) , a+2c=1, a=(1/3) , b=0  ⇒((1+sh^2 u)/(shu(3+2sh^2 u)))=(((1/3)shu)/(3+2sh^2 u))+((1/3)/(shu))  ⇒∫((ch^2 u)/(shu(3+2sh^2 u)))du=(1/3){∫((shu du)/(3+2ch^2 −2))+∫(du/(shu))}  =(1/3){(1/(√2))∫((d((√2)chu))/(1+2ch^2 u))+∫((shu du)/(ch^2 u−1))}  =(1/3){(1/(√2))Arctan((√2)chu)−Arctanh(chu)}