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. 

Note that in the particular case of $\sin^2(x)$ and $\cos^2(x)$ there is a trick you can use.
$\begin{cases}\cos^2(x)+\sin^2(x)=1\\\cos^2(x)-\sin^2(x) = \big(\sin(x)\cos(x)\big)' \end{cases}$
Thus if we call $C=\int\cos^2$ and $S=\int\sin^2$ we get the system below
$\begin{cases}C(x)+S(x)=x+cst\\C(x)-S(x)=\sin(x)\cos(x)+cst\end{cases}$
Which easily solves to
$\begin{cases}C(x)=\frac x2+\frac 12\sin(x)\cos(x)+cst\\S(x)=\frac x2-\frac 12\sin(x)\cos(x)+cst\end{cases}$
In your problem, you just translate by $\frac\pi6$ and get $I = \frac{x+\frac\pi6}2-\frac 12\sin(x+\frac\pi6)\cos(x+\frac\pi6)+cst$
I let convince yourself it is the same answer than Atmos's ($\frac{\pi}{12}$ is absorbed by the $cst$ term, and $\sin\cos$ is transformed using the doubled angle).
