Today, we continue to look at integrals using the different integration techniques shown in Calculus 2. In today’s example we find \(\int \sin^{2}(x)\cos^{2}(x)dx\). After determining that the best technique of integration to do this is by using reduction formulas, we then evaluate this integral.
Hopefully you will join me each day as I go through these integrals in both a post and in a video. Along the way, I hope you learn something, and, if you are taking Calculus 2, you are in a great position to do very well on any upcoming exams you may have. You can find more Calculus 2 study help posts here. I also have a decision tree available so that you can follow along as we determine what technique of integration to use. I also have a useful list of formula sheet available.
 Can you guess the answer.
Since it is unlikely we will be able to guess the answer, we will say no and move onto the next question.  Is there an inside function?
At this point, there is indeed an inside function. In fact, we have \(i=\cos(x)\) or \(i=\sin(x)\) as options. We will therefore choose one, so let \(i=\sin(x)\). Then \(di=\cos(x)dx\) and \(dx=\frac{di}{\cos(x)}\). Using this, we now find that
\begin{align*}
\int \cos^{2}(x)\sin^{2}(x)dx &=\int \cos^{2}(x)i^{2}\frac{di}{\cos(x)} \\
&=\int \cos(x)i^{2}di.
\end{align*}  Are all the \(x\)s gone?
No, we were not able to cancel out all the \(x\)s.  Do the remaining terms with \(x\)s look like \(i\)?
While there are similarities between \(\cos(x)\) and \(\sin(x)\), we do not have that \(\cos(x) \neq \sin(x)+c\) for any constant \(c\). Therefore, we start over using different decisions along the way.
Second Time Through
We begin again by looking at our decision tree. We quickly note that we cannot guess the answer, because we couldn’t before, so we move onto the next question. In this case, there is still an inside function we haven’t tried, that is, we could pick \(i=\cos(x)\). I would suggest looking at this, however, if you do, you will note that your result is extremely similar to the one we got using \(i=\sin(x)\) except we will have a \(\sin(x)\) left over this time that we cannot cancel out. We
We will therefore move past this questions and arrive at

 Is there a product of functions?
Yes, we do have the product of functions.
 Is there a product of functions?

 Is it the product of trig functions?
Here, we do indeed have a product of trig functions. This tell us that we want to try to use reduction formulas.
 Is it the product of trig functions?
 When using reduction formulas, we first want to simplify the problem by writing in terms of one trig function. In order to do this, we note either that \(\cos^{2}(x)=1\sin^{2}(x)\) or \(\sin^{2}(x)=1\cos^{2}(x)\). Either choice is fine here, so I will use the first. We then get that
\begin{align*}
\int \cos^{2}(x)\sin^{2}(x)dx&=\int (1\sin^{2}(x))\sin^{2}(x)dx \\
&=\int \sin^{2}(x)\sin^{4}(x)dx \\
&=\int \sin^{2}(x)dx \int \sin^{4}(x)dx.
\end{align*}  In order to continue with this problem, we need the reduction formula \(\int \sin^{n}(x)dx=\frac{cos(x)\sin^{n1}(x)(n1)\int sin^{n2}(x)dx}{n}+c\). We then get that
 \begin{align*}
\int \cos^{2}(x)\sin^{2}(x)dx&=\int (1\sin^{2}(x))\sin^{2}(x)dx \\
&=\int \sin^{2}(x)\sin^{4}(x)dx \\
&=\int \sin^{2}(x)dx \int \sin^{4}(x)dx \\
&=\frac{\cos(x)\sin(x)+\int 1 dx}{2}\frac{\cos(x)\sin^{3}(x)3\int \sin^{2}(x)dx}{4} \\
&=\frac{\cos(x)\sin(x)+x}{2}\frac{\cos(x)\sin^{3}(x)3\left(\frac{\cos(x)\sin(x)+x}{2}\right)}{4}+c \\
&=\frac{\cos(x)\sin^{3}(x)}{4}\frac{\cos(x)\sin(x)}{8}+\frac{x}{8}+c.
\end{align*}  Now that we have arrived at the answer we should check our work. That is, we should take the derivative of the answer and make sure that it coincides with our original integrand. I will leave this process to you, but if you need help, please check out our posts on derivatives also available in the Calculus posts. We will indeed get our original if properly differentiate and simplify. Therefore, we have found are integral.

Conclusion
 In this problem, our initial trial did not work out. There is nothing wrong with that, as part of the process of learning is making (many) mistakes. However, we recognized that the technique we originally chose wouldn’t work, so, instead of giving up, we tried something else. We were then able to use our reduction formulas in order to evaluate our integral.
 I hope you learned something from this post and found it helpful. If you did, make sure to share it and the video with anyone else you know that may need some extra help with integrals.
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