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February 18, 2026
Section 7.2 : Integrals Involving Trig Functions
3. Evaluate \( \displaystyle \int{{{{\cos }^4}\left( {2t} \right)\,dt}}\).
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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 2Now 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 3We’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 4Now all we have to do is evaluate the integral.
\[\int{{{{\cos }^4}\left( {2t} \right)\,dt}} = \frac{1}{4}\left( {\frac{3}{2}t + \frac{1}{2}\sin \left( {4t} \right) + \frac{1}{{16}}\sin \left( {8t} \right)} \right) + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{8}t + \frac{1}{8}\sin \left( {4t} \right) + \frac{1}{{64}}\sin \left( {8t} \right) + c}}\]