# Understanding application of trig identity $\cos^2\theta = \frac{1+\cos2\theta}{2}$ in integration

I'm in Calc II and am (still) rusty on my trig after not having taken math for over a decade. I'm trying to understand the application of the trig identity below.

\begin{align} \int^{\pi}_0 \cos^4 (2t) dt &= \int^{\pi}_0 (\cos^2 (2t))^2 dt\newline \color{red}{\text{Using the property } \cos^2 \theta = \frac{1 + \cos 2\theta}{2}}\newline &= \int^{\pi}_0 \left(\frac{1 + \color{blue}{\cos(4t)}}{2}\right)^2 dt \end{align}

The highlighted portion is the part I'm struggling with. Because it's the $\cos^2(2t)$, rather than $\cos^2(t)$, is the idea that we let $\theta = 2t$, and so $2\theta = 4t$? Would this apply to any input into the trig function? For example, if it was $\cos^2(\frac t2 + 1)$, could I let $\theta = \frac t2 + 1$ and $2\theta = t + 2$, so that: $$\cos^2(\frac t2 + 1) = \frac{1 +\cos(t + 2)}{2} ?$$

Confirmation that I'm understanding this correctly would be helpful.

• Yes, you understood correctly. Commented Sep 9, 2017 at 18:23
• The next step would be expanding the square and using the same technique to deal with $\cos^2(4t)$ in the numerator. Should be a fairly easy integral then! Commented Sep 9, 2017 at 18:35

You are absolutely correct. You can also attempt to use integration by parts, but the $\cos^2 t = (1 + \cos 2t)/2$ substitution saves a lot of grief.