Calculus 3..Evaluating line integral

Evaluate the line integral, where C is the given curve.
∫C x sin(y)ds, C is the line segment from (0, 1) to (4, 4)


I want to know what is x and y?? I got x =4t and y=3t+1. But after solving it I am not getting the correct result.

You obtained x and y correctly (which is valid on [0, 1]).

Note that
ds = sqrt[(dx/dt)^2 + (dy/dt)^2] dt = sqrt(4^2 + 3^2) dt = 5 dt.

Therefore, ∫c x sin y ds
= ∫(t = 0 to 1) 4t sin(3t + 1) * 5 dt
= ∫(t = 0 to 1) 20t sin(3t + 1) dt
= 20t * -cos(3t + 1)/3 {for t = 0 to 1} - ∫(t = 0 to 1) 20 * (-cos(3t + 1)/3) dt
= [(-20t/3) cos(3t + 1) + 20 sin(3t + 1)/9] {for t = 0 to 1}
= (-20/3) cos(4) + 20 sin(4)/9 - 20 sin(1)/9.

I hope this helps!

Using your parameterization, did you forget that ds = sqrt( dx^2+dy^2) dt which needs to be included in the integral. Your integral should be
int(4t sin(3t+1) * sqrt(16+9) dt) = 5 int(4t sin(3t+1) dt) = 20/9 [ sin(3t+1) - 3t cos(3t+1)]
and plug in your limits of integration. Hope that helps.