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

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:
${\displaystyle \gamma \left(t\right)=(0,3)+(4t,3t)=(4t,3t+3),t\in [0,1]}$
Next, find the derivative of the parametrization.
${\displaystyle {\gamma }^{\prime }\left(t\right)=(4,3)\Rightarrow \left|\left|{\gamma }^{\prime }\left(t\right)\right|\right|=\sqrt{4^2+3^2}=5}$
Finally, we calculate the integral:
${\displaystyle {\int }_0^{1}4t\sin (3+3t)\cdot 5dt=20{\int }_0^{1}t\sin (3+3t)dt=\frac{20\sin 6-60\cos 6-20\sin 3}{9}}$
Result:
${\displaystyle \frac{20\sin 6-60\cos 6-20\sin 3}{9}}$