To answer this question, split up the line integral into two pieces:

intc (x + 2y)dx and intc (x - y)dy.

Our parameter is t, 0<=t<=pi/4 (I assume, because your problem statement gives inf <-- t < 0, which diverges )

We need to convert dx, dy into dt:

...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

To answer this question, split up the line integral into two pieces:

intc (x + 2y)dx and intc (x - y)dy.

Our parameter is t, 0<=t<=pi/4 (I assume, because your problem statement gives inf <-- t < 0, which diverges )

We need to convert dx, dy into dt:

x = 2 cos t --> dx = -2 sin t dt

y = 4 sin t --> dy = 4 cos t

Now,

intc (x + 2y)dx = int [ (2 cos t + 8 sin t) ( -2 sin t ) dt , 0<= t <=pi/4]

= int [ -4 costsint - 16sint^2, 0<= t <=pi/4 ]

And,

= intc (x - y)dy = int [ (2 cos t - 4 sin t) ( 4 cos t ) dt , 0<= t <=pi/4]

= int [ 8 cost^2 - 16 costsint, 0<= t <=pi/4 ]

So,

intc (x + 2y)dx + intc (x - y)dy = int[ 8 cost^2 - 20 costsint - 16 sint^2, 0<= t <=pi/4 ]

=-4 t + 5 cos(2 t) + 6 sin(2 t) + C, evaluated from 0<=t<=pi/4

**= 1 - pi = 2.14**