How can I find $\int_0^{2\pi}\sin(x)\sin(x+1)$

Question

I'm trying to find $\int_0^{2\pi}\sin(x)\sin(x+1)$, however, I'm having a lot of trouble.

I've tried using integration by parts on it, but when I ended up with $\int \sin(x)\sin(x+1)$ on both the left and right sides, they just cancelled each other out.

I also tried using the trig identity $\sin A\sin B = \frac12 \cos(A-B) - \frac12 \cos(A+B)$, but when I tried to integrate that to reach an answer, I'm apparently doing it wrong somewhere. For reference, the final indefinite integral I reached was:

$\frac12\sin(-1) - \frac14\sin(2x+1)$.

I was hoping someone would know where I went wrong here. Thanks very much!

Answer 1

Hint

As you said, you can write

$$\sin x\sin (x+1)=\frac12 \cos (1)-\frac12 \cos(2x+1)$$

See that $\cos (1)$ is a constant number. Now, you have to solve

$$\int_{0}^{2\pi}\cos(1)dx=\cos(1)\int_{0}^{2\pi}1dx$$ $$\int_{0}^{2\pi}\cos(2x+1)dx.$$

Can you finish?

Answer 2

$$sin{(x)}\sin{(x+1)}=\frac{1}{2}(\cos{1}-\cos{(2x+1)})$$ $$\displaystyle \int_{0}^{2\pi}\sin{(x)}\sin{(x+1)}dx=\pi\cos{1}-\frac{sin(4\pi+1)}{4}+\frac{\sin{1}}{4}$$