complex analysis - Calculating a contour integral

I want to evaluate the integral

$$\int_{\gamma} \sin{(2z)} \ {\rm d}z$$ where $\gamma$ is the line segment joining the point $i+1$ to the point $-i$.

Thus $\gamma(t) = -i+t(2i+1)$ for $0\le t\le1$.

So I want to calculate

$$\begin{align}\int_{\gamma} \sin{(2z)} \ {\rm d}z &=\int^{1}_{0} f(\gamma(t))\gamma'(t) \ {\rm d}t\\ &=\int_{0}^{1} \sin{[2(-i+t(1+2i))]}(1+2i) \ {\rm d}t \\ &=(1+2i)\int^{1}_{0} \sin{[2t+i(4t-2)]} \ {\rm d}t \\ &= (1+2i)\int^{1}_{0} \sin{(2t)}\cosh{(2-4t)}-i\cos{(2t)}\sinh{(2-4t)} \ {\rm d}t \\ &=(1+2i)\left[\int^{1}_{0}\sin{(2t)}\cosh{(2-4t)} \ {\rm d}t\, - i\int^{1}_{0}\cos{(2t)\sinh{(2-4t) \ {\rm d}t}}\right]\end{align}$$

Now this seems extremely long winded, is there any other way to calculate this?

Answer

We have $\gamma(t)=-i+t(2i+1)$ for $0\le t\le 1$. Since $\gamma$ is smooth and $f(z) = \sin{(2z)}$ is continuous, let $F = \int f$ and note $\gamma(1)=1+i$, $\gamma(0)=-i$. By the fundamental theorem of calculus applied to contour integrals

$$\int_{\gamma} f = F(\gamma(1))-F(\gamma(0)).$$

Therefore

$$\begin{align}\int_{\gamma} \sin{(2z)} \ {\rm d}z &= -\frac{1}{2}\cos{(2(1+i))}+\frac{1}{2}\cos{(2(-i))} \\ &= \frac{1}{2}\left[\cos{(2i)-\cos{(2+2i)}}\right].\end{align}$$