Let $\gamma$ be a upper half of unit circle. Prove: $$\left | \int_{\gamma} \dfrac{\sin z}{z+i}dz \right | \leq \dfrac{\pi \sinh 1}{\sqrt{2}}$$
My try:
Using Jordan lemma, I get : $\left | \int_{\gamma} \dfrac{e^{iz}}{z+i}dz \right | \leq \dfrac{\pi }{\sqrt{2}}$. And if R big enough $\left | \int_{C_R} \dfrac{\sin z}{z+i}dz \right |= \dfrac{\pi \sinh 1}{2}$ .
If $z$ belongs to the upper half of unit circle, then $\lvert z+i\rvert\geqslant\sqrt2$ and\begin{align}\bigl\lvert\sin(z)\bigr\rvert&=\left\lvert z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots\right\rvert\\&\leqslant\lvert z\rvert+\frac{\lvert z\rvert^3}{3!}+\frac{\lvert z\rvert^5}{5!}+\cdots\\&\leqslant1+\frac1{3!}+\frac1{5!}+\cdots\\&=\sinh(1).\end{align}Since the length of the upper half of unit circle is $\pi$, you get that$$\left\lvert\int_\gamma\frac{\sin z}{z+i}\,\mathrm dz\right\rvert\leqslant\frac{\pi\sinh(1)}{\sqrt2}$$indeed.