integration question no.5
Use u = sinT?
[Let me write T as theta---for less typing.]
Okay.
u = sinT
So,
du = cosT dT
dT = du/cosT
INT.[sin(2T) / cbrt(sin^2(T) +3)]dT
= INT.[2sinTcosT *(sin^2(T) +3)^(-1/3)]dT
Substitutions,
= INT.[2u*cosT*(u^2 +3)^(-1/3)](du/cosT)
= INT.[2u*(u^2 +3)^(-1/3)]du
= INT.[(u^2 +3)^(-1/3)](2u du)
d(u^2 +3) = 2u du
= INT.[(u^2 +3)^(-1/3)]d(u^2 +3)
= 1/(-1/3 +1) * (u^2 +3)^(-1/3 +1) +C
= (3/2)(u^2 +3)^(2/3) +C
Going back to u = sinT
= (3/2)(sin^2(T) +3)^(2/3) +C -----------------answer.