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# Evaluating line integral directly - part 2

Finishing up the line integral with a little trigonometric integration. Created by Sal Khan.

## Want to join the conversation?

• At around , can I replace "4 cos θ sin θ" with "2 sin 2θ" so that later on my integral would become "-cos 2θ" instead of 2 sin^2 θ"?
• Yes, you can. 2sin²(θ) is the same as 1 - cos(2θ), and that 1 is just a constant that can be ignored, because it is the same when evaluated at 0 and at 2π.
• could we have taken c as the upper slanted curve?
• The integral was made using the curve `c`. The parametrisation shows this, the `r⃗(θ)` used was: `r⃗(θ) = cos(θ)î + sin(θ)ĵ + (2 - sin(θ))k̂` The x and y coordinates are the unit circle (seen from above) and the z-coordinate is the "slant".
• If this curve was 3D do we have to dot 3d force field vector with normal unit vector to the surface first?
• how does 1/2cos(2x) = d/dx (1/4sin(2x) *substituting x for theta
(1 vote)
• Because d/dx(sin(2x)) = 2*cos(2x) (by Chain rule).
• I understand r = 1 in this case, but in other cases, do we need to multiply by an integrating factor?
(1 vote)
• How do you know when to use Stokes' Theorem vs directly integrating it?
(1 vote)
• I feel like this method is easier than stokes theorem