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

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!

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?

• Yeah, I managed to finish it! I forgot to take into account that the cos(1) was a constant, so it could be moved out. Thank you very much! Commented Mar 9, 2021 at 17:23

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