Finding limits of line integral

[boneill3]

Homework Statement

Integrate along the line segment from (0,0) to (\pi,-1)
The integral

\int_{(0,1)}^{(\pi,-1)} [y sin(x) dx - (cos(x))]dy

Homework Equations

The Attempt at a Solution

I have used the parameterization of x=\pi t and y= 1-2t
To get the integral:
\int_{(0,1)}^{(\pi,-1)} [1-2t sin(\pi t) -(cos(\pi t))]dt

But now because it is an integral of variable t I need to change the limits .

I'm not sure if I just have to put the limits of t just from 0 to \pi

I suppose I'm having trouble with getting from the limit of 2 variables (x,y) to a limit of one variable t

Thanks

 

[Dick]

If x=pi*t and y=1-2t, then if you put t=0 then x=0 and y=1, right? If you put t=1 then x=pi and y=(-1), also right? As you came up with that fine parametrization what's the problem with finding limits for t?

 

[boneill3]

Thank you for your help.
I will need to go back and study more about parametrization.

 

[boneill3]

When calculating this line integral

\int_{(0,1)}^{(\pi,-1)} [y sin(x) dx - (cos(x))]dy

I'm using the formula
\int_{a}^{b}[f(x(t),y(t))x'(t) + g(x(t),y(t))y'(t)]dt

with parameterization

I have x = \pi t
y = 1-2t
so
x' = \pi
and
y' = -2

plugging into the integral I get

\int_{(0)}^{(1)} [1-2y sin(\pi t) \pi - (cos(\pi t))-2]
= -1

The question states that the integral is independant of path.

So if I integrate along the initial line segment (0,1)to (0,\pi)
I should be able to plug in the values f((-1,\pi))-f((0,1))
And it should equal my original integral vaue of -1.

However I get 0

Could someone please check what I've done I show me where I am going wrong ?