$$v(t) = 3 - 2\sin(t) + 8\sin^2(t)$$
To find the rms of this function, I first figured out that the period $T = 2\pi$. I then set up the equation:
$V = \sqrt{\frac{1}{T}\int^T_0v^2(t)\,dt}$
Calculating $v^2(t)$ is easy, for me, however I thought the process of integrating $v^2(t)$ seemed very tedious to me without using a computer.
Am I using the best (easiest in this case) method for figuring out the root mean square for my function $v(t)$?
Using $$v(t) = 3 - 2\sin(t) + 8\sin^2(t)$$ and squaring (just brute force), you have (just set $x=\sin(t)$ and use binomial expansion)$$v^2(t)=9-12 \sin (t)+52 \sin ^2(t)-32 \sin ^3(t)+64 \sin ^4(t)$$ Now, you can use power reduction formulae which give $$\sin^2(x)=\frac{1-\cos(2x)}2$$ $$\sin^3(x)=\frac{3\sin(x)-\sin(3x)}4$$ $$\sin^4(x)=\frac{3-4\cos(2x)+\cos(4x)}8$$ Replacing, you the have $$v^2(t)=59-36 \sin (t)-58 \cos (2 t)+8 \sin (3 t)+8 \cos (4 t)$$