How to Simplify a Curve Integration Problem Using Trigonometric Identities

[Rhduke]

Homework Statement

Well i got the formula down, I am having trouble with the integration though:
Q: {(x^3 + y) ds , x = 3t, y = t^3, 0<=t<=1

= {[ (27t^3 + t^3) sqrt(9t^2 + t^6)] dt

Homework Equations

f(x,y)ds = f(x(t), y(t)) sqrt[ (x'(t))^2 + (y'(t))^2 ]

The Attempt at a Solution

I simplified it down to: 28 {[ (t^4) sqrt(9 + t^4) ] dtAny help would be appreciated.

t^2 = 3cosu
= 28{ [(9cos^2u) sqrt(9 + 9cos^2u)] du
=756{ [(cos^2u) sqrt(1 + cos^2u)] du
no helpful identity..
dt^2 = -3sinu du

 

[HallsofIvy]

WHY would you let t^2= 3 cos u? Since sin^2(x)+ cos^2(x)= 1, you have sin^2(x)= 1- cos^2(x) so "v= cos(x)" is useful for things like 1- cos^2(x), not 1+ cos^2(x).

Rather, dividing sin^2(x)+ cos^2(x)= 1 by cos^2(x) you have tan^2(x)+ 1= sec^2(x) so t^2= 3tan(u) would be much better.