Why does $\int^\limits{1}_0\cos(3x+5)dx$ not have an indefinite integral?

Question

I recently came across a book introducing Simpson's rule and how it can be used to approximate the area under a curve that cannot be integrated symbolically. The book gave $\int^\limits{1}_0\cos(3x+5)dx$ as an example of such a function, however I don't understand why this function cannot be integrated symbolically; upon trying myself I found its integral to be $\frac{1}{3}\sin(3x +5) + c$. Am I missing something here?

Answer 1

It can.

The integral is trivial: $$y = 3x + 5$$ $$\text{d}x = \frac{1}{3}\ \text{d}y$$

$$\frac{1}{3} \int_5^8\cos(y) \ \text{d}y = \frac{1}{3} (\sin (8)-\sin (5))$$

And in general

$$\int_a^b \cos(y)\ \text{d}y = \sin(b) - \sin(a) + C$$

The Simpson way is just to make you exercise on numerical calculations.

Id est: to see how much good is Simpson rule to approximate the true result of that integration (read above).