the integral from 0 to pi/2 of acost *bsint *sqrt(a^2*sin^2 t + b^2*cos^2 t)dt
Thank you very much.
INT.(0-->pi/2)[(a*sinT)(b*cosT)sqrt[(a^2)sin^2(T) +(b^2)cos^2(T)]]dT
= INT.(0-->pi/2)[sqrt[(a^2)sin^2(T) +(b^2)cos^2(T)]*[ab*sinTcosT dT] -----(i)
(a^2)sin^2(T) +(b^2)cos^2(T)
= (a^2)(1 -cos^2(T)) +(b^2)cos^2(T)
= a^2 -(a^2)cos^2(T) +(b^2)cos^2(T)
= a^2 -(a^2 -b^2)cos^2(T)
And the derivative of that is
d/dt [a^2 -(a^2 -b^2)cos^2(T)]
= 0 -(a^2 -b^2)*(2cosT)(-sinT dT)
= 2(a^2 -b^2)sinTcosT dT
Use that in (i),
INT.(0-->pi/2)[sqrt{(a^2)sin^2(T) +(b^2)cos^2(T)}]*[ab*sinTcosT dT]
= INT.(0-->pi/2)[sqrt{(a^2)sin^2(T) +(b^2)cos^2(T)}]*[ab*sinTcosT dT]*[{2(a^2 -b^2)/ab} /
{2(a^2 -b^2)/ab}]
= [ab/ 2(a^2 -b^2)]*INT.(0-->pi/2)[sqrt{(a^2)sin^2(T) +(b^2)cos^2(T)}]*[2(a^2 -b^2)sinTcosT
dT]
That is in the form
INT.[u^(1/2)]du = (2/3)U^(3/2) +C
So,
= [ab/ 2(a^2 -b^2)](2/3){(a^2)sin^2(T) +(b^2)cos^2(T)}*(3/2) | (0-->pi/2)
= [ab/ 3(a^2 -b^2)]{(a^2)sin^2(T) +(b^2)cos^2(T)}*(3/2) | (0-->pi/2)
= [ab/ 3(a^2 -b^2)]*[{(a^2)sin^2(pi/2) +(b^2)cos^2(pi/2)}^(3/2) -{(a^2)sin^2(0)
+(b^2)cos^2(0)}^(3/2)}]
= [ab/ 3(a^2 -b^2)]*[{(a^2)(1) +(b^2)(0)}^(3/2) -{(a^2)(0) +(b^2)(1)}^(3/2)}]
= [ab/ 3(a^2 -b^2)]*[(a^2)^(3/2) -(b^2)^(3/2)]
= [ab/ 3(a^2 -b^2)]*[a^3 -b^3]
= (1/3)[ab(a^3 -b^3) / (a^2 -b^2)]
= (1/3)[ab(a-b)(a^2 +ab +b^2) / (a-b)(a+b)]
= (1/3)[ab(a^2 +ab +b^2) / (a+b)] ---------------------answer.