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

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?

• Maybe the purpose is to check the approximation. – Olivier Oloa Nov 20 '16 at 19:13
• @OlivierOloa I would have thought so myself, however the book explicitly states that cos(3x + 5) may not be integrated symbolically – DividedByZero Nov 20 '16 at 19:25

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).

• Of course you can evaluate it even without the substitution, that was just to make use of it. Simpson rule will approximate the solution. In that case is quite useless, but it works good in other cases, in which you cannot get a primitive. – Von Neumann Nov 20 '16 at 19:16
• @RandomUser If someone (your professor or the book) told you that it's not possible to evaluate it, then yes it's an error. – Von Neumann Nov 20 '16 at 19:17
• Yeah, the book said that one simply cannot evaluate said function's indefinite integral at all.. Anyway, thanks for the answer! – DividedByZero Nov 20 '16 at 19:19
• @RandomUser You're welcome! – Von Neumann Nov 20 '16 at 19:22