# Evaluate the line integral, where C is the given curve. Integral C x \sin

Evaluate the line integral, where C is the given curve. Integral C x $\mathrm{sin}$ y ds, C is the line segment from (0, 3) to (4, 6)
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bahaistag
Recall the way we parametrize the segment between the two points. First, find the vector determined by the given two points:
(4,6)-(0,3)=(4,3)
The starting point is (0,3). The segment can be now parametrized as:
$\gamma \left(t\right)=\left(0,3\right)+\left(4t,3t\right)=\left(4t,3t+3\right),t\in \left[0,1\right]$
Next, find the derivative of the parametrization.
${\gamma }^{\prime }\left(t\right)=\left(4,3\right)⇒||{\gamma }^{\prime }\left(t\right)||=\sqrt{{4}^{2}+{3}^{2}}=5$
Finally, we calculate the integral:
${\int }_{0}^{1}4t\mathrm{sin}\left(3+3t\right)\cdot 5dt=20{\int }_{0}^{1}t\mathrm{sin}\left(3+3t\right)dt=\frac{20\mathrm{sin}6-60\mathrm{cos}6-20\mathrm{sin}3}{9}$
Result:
$\frac{20\mathrm{sin}6-60\mathrm{cos}6-20\mathrm{sin}3}{9}$