# Can someone just check if I did this line integral correctly?

## The Attempt at a Solution

$$\int _{ 0 }^{ 2\pi }{ xdx } \\ =-\int _{ 0 }^{ 2\pi }{ sintcostdt } \\ =0$$

It feels wrong.

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I got zero also using a different method. Let's review the steps. You said
x = cos t, dx = -sint dt and t now runs from 0 to 2##\pi##.

Since cost and sint are both periodic in 2##\pi## you got zero.

What may feel wrong is that ##\int_0^{2\pi}x dx## is not zero. However, this is not the same thing as integrating x around a circle. Basically you are running cos(t) around that circle, and it's going to start and end with the same value.

HallsofIvy