Find $\int_C v\cdot t ds$ when $C$ is $y=\sin x, 0\leq x \leq \pi $ and $v(x,y) = (y+e^x, 1)$, and $t$ is the unit tangent to $C$.
This is the first question of a vector analysis exam I am doing and I keep getting something not very nice, and I am not sure if the lecturer has made a mistake or if I am wrong.
So I parameterise $C$ by $r(t) = (t, \sin t), 0 \leq t \leq \pi$.
The unit tangent to $C$ is then $\frac{r'(t)}{||r'(t)||} = \frac{1}{\sqrt{1+\cos^2 t}} (1, \cos t)$
So then $$\int_C v\cdot t ds = \int_0^{\pi} (\sin t + e^t, 1) \cdot \left(\frac{1}{\sqrt{1+\cos^2 t}} (1, \cos t)\right)dt$$ $$ = \int_0^{\pi}\frac{1}{\sqrt{1+\cos^2 t}} (\sin t + e^t + \cos t)dt$$
which I see no easy way of integrating.
I presume I am wrong somewhere, but I don't know where. A picture of the actual question is attached if my recreation isn't clear enough. 
You forgot your $ds$ in $$ \int_0^{\pi}\frac{1}{\sqrt{1+\cos^2 t}} (\sin t + e^t + \cos t)dt$$
When you include your ds, the integral simplifies and you can evaluate it easily.