Show that $\int_0^a \frac{\sin x}{x} dx \leq 1+\log a$ for all $a \geq 1$

It looks good to me, one other way to approach this problem is $$ \int_0^a \frac{\sin(x)}{x}dx=\int_0^1\frac{\sin(x)}{x}dx+\int_1^a\frac{\sin(x)}{x}dx $$ But as you said $\displaystyle\int_0^1 \frac{\sin(x)}{x}dx\leqslant 1$ because $\sin(x)\leqslant x$, and on the other hand $$ \int_1^a\frac{\sin(x)}{x}dx\leqslant \int_1^a\frac{dx}{x}=\log(a)$$ Therefore $$ \int_0^a \frac{\sin(x)}{x}dx\leqslant 1+\log(a) $$