1. ## integration - urgent help please

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.

2. hi turkeywilliam,

why ? Could you please explain to me for this link. thank you .

also, i have the final answer, given by my teacher, is [ab(a^2 +ab+b^2) ]/3(a+b).

3. $ab \int_{0}^{\pi/2} \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt$

Let $x = a \cos t$ and $y = b \sin t$ (elliptic integral).

4. 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.