<s>[INST] Evaluate integral for $\int \sin^2 (x+\frac{\pi}{6}) dx$ Can someone walk me through how to evaluate the integral $$\int \sin^2 (x+\frac{\pi}{6}) dx?$$
I get as far as
$$\int \frac {1 - \cos(x + \frac{\pi}{6})}2dx,$$
but I am not sure how to proceed. 
 [/INST] Careful with formula
$$
\sin^2\left(t\right)=\frac{1-\cos\left(2t\right)}{2}
$$
Then using linearity
$$
\int \frac{1-\cos\left(2x+\frac{\pi}{3}\right)}{2}\text{d}x=\int \frac{1}{2}-\frac{1}{2}\int \cos\left(2x+\frac{\pi}{3}\right)
$$
Hence

$$
\int \frac{1-\displaystyle \cos\left(2x+\frac{\pi}{3}\right)}{2}\text{d}x=\frac{x}{2}-\frac{1}{4}\sin\left(2x+\frac{\pi}{3}\right)+K$$

</s>