Section 7.2 : Integrals Involving Trig Functions
3. Evaluate \( \displaystyle \int{{{{\cos }^4}\left( {2t} \right)\,dt}}\).
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Hint : Pay attention to the exponents and recall that for most of these kinds of problems you’ll need to use trig identities to put the integral into a form that allows you to do the integral (usually with a Calc I substitution).
Start Solution
The first thing to notice here is that we only have even exponents and so we’ll need to use half-angle and double-angle formulas to reduce this integral into one that we can do.
Also, do not get excited about the fact that we don’t have any sines in the integrand. Sometimes we will not have both trig functions in the integrand. That doesn’t mean that that we can’t use the same techniques that we used in this section.
So, let’s start this problem off as follows.
\[ \int{{{{\cos }^4}\left( {2t} \right)\,dt}} = \int{{{{\left( {{{\cos }^2}\left( {2t} \right)} \right)}^2}\,dt}}\]Show Step 2
Now we can use the half-angle formula to get,
\[\int{{{{\cos }^4}\left( {2t} \right)\,dt}} = \int{{{{\left[ {\frac{1}{2}\left( {1 + \cos \left( {4t} \right)} \right)} \right]}^2}\,dt}} = \int{{\frac{1}{4}\left( {1 + 2\cos \left( {4t} \right) + {{\cos }^2}\left( {4t} \right)} \right)\,dt}}\]Show Step 3
We’ll need to use the half-angle formula one more time on the third term to get,
\[\begin{align*}\int{{{{\cos }^4}\left( {2t} \right)\,dt}} & = \frac{1}{4}\int{{1 + 2\cos \left( {4t} \right) + \frac{1}{2}\left[ {1 + \cos \left( {8t} \right)} \right]\,dt}}\\ & = \frac{1}{4}\int{{\frac{3}{2} + 2\cos \left( {4t} \right) + \frac{1}{2}\cos \left( {8t} \right)\,dt}}\end{align*}\]Show Step 4
Now all we have to do is evaluate the integral.
\[\int{{{{\cos }^4}\left( {2t} \right)\,dt}} = \frac{1}{4}\left( {\frac{3}{2}t + \frac{1}{2}\sin \left( {4t} \right) + \frac{1}{{16}}\sin \left( {8t} \right)} \right) + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{8}t + \frac{1}{8}\sin \left( {4t} \right) + \frac{1}{{64}}\sin \left( {8t} \right) + c}}\]