In Example 5.6.3, we performed a \(u\)-substitution, but needed to work to rewrite our whole integrand in terms of \(u\text{.}\) Specifically, we found that in the numerator, there was an \(x^3\text{,}\) but \(du=2x\;dx\text{.}\) We were substituting out a linear function of \(x\) in the numerator, but the actual function was cubic. This wasn’t a problem: we rewrote \(x^3=x^2\cdot x\text{,}\) and noticed that the extra \(x^2\) was able to be substituted, since we could rewrite our substitution rule: we noted that \(u=x^2+1\) is equivalent to \(x^2=u-1\text{.}\) This meant that even though we had an extra factor of \(x^2\) “in” the part that we were using for substituting in the differential \(du\text{,}\) we were still able to translate the whole function to be written in terms of \(u\text{.}\)
On one hand, we can look at the structure of the integrand and notice that we have a product of two functions! Integration by parts was a fine strategy to employ, and that’s what we did in the example. On the other hand, we noticed that since we have this function-derivative pairing, a \(u\)-substitution also was appropriate.
In this section, we’ll explore more combinations of trigonometric functions and build a strategy for antidifferentiating them that includes some ideas from both of these previous examples.
Consider the first integral,\(\displaystyle\int\sin^4(x)\cos(x)\;dx\text{.}\) Think about and set up a good technique for antidifferentiating. Without actually solving the integral, explain why this technique will work.
It might be helpful to notice that \(\sin^4(x)\) can be rewritten as \(\left(\sin(x)\right)^4\text{.}\) Does this help reveal something important about the structure of this integrand?
Now consider the second integral,\(\displaystyle\int\sin^4(x)\cos^3(x)\;dx\text{.}\) Does the same integration strategy work here? What happens when you apply the same thing?
We know that \(\sin(x)\) and \(\cos(x)\) are related to each other through derivatives (each is the derivative of the other, up to a negative). Is there some other connection that we have between these functions? We might especially notice that we have a \(\cos^2(x)\) left over in our integral. Can we write this in terms of \(\sin(x)\text{,}\) so that we can write it in terms of \(u\text{?}\)
Why would this strategy not have worked if we were looking at the integrals \(\int\sin^4(x)\cos^2(x)\;dx\) or\(\displaystyle\int\sin^4(x)\cos^4(x)\;dx\text{?}\) What, specifically, did we need in order to use this combination of substitution and trigonometric identity to solve the integral?
For integrals in the form\(\displaystyle\int\sin^p(x)\cos^q(x)\;dx\) where \(p\) and \(q\) are real number exponents:
If \(q\text{,}\) the exponent on \(\cos(x)\) is odd, we should use \(u=\sin(x)\) and \(du=\cos(x)\;dx\text{.}\) Then, we can apply the Pythagorean Identity \(\cos^2(x)=1-\sin^2(x)\text{.}\)
If \(p\text{,}\) the exponent on \(\sin(x)\) is odd, we should use \(u=\cos(x)\) and \(du=-\sin(x)\;dx\text{.}\) Then, we can apply the Pythagorean Identity \(\sin^2(x)=1-\cos^2(x)\text{.}\)
A strange note, here, is that we typically pick our \(u\)-substitution based on looking to see a suitable candidate for \(u\text{:}\) we look for functions that are composed “inside” of other functions or we look for a function whose derivative is in the integral (the “function-derivative pair” that we talk about in Section 5.6). Here, though, we’re selecting our substitution based on \(du\text{:}\) we’re looking to see which function we can set aside one copy of for the differential, and then have an even power left over so that we can apply the Pythagorean Identity to translate the rest.
Notice that the exponent on \(\sin(x)\) is odd: if you let \(u=\cos(x)\text{,}\) you’ll end up with \(\sin^4(x)\) left over in your integral, and you can write it as \((\underbrace{1-\cos^2(x)}_{\sin^2(x)})^2\text{.}\)
This one seems scary at first, because of the fraction exponent. Notice, though, that you have no hope of converting fractional exponents of sine functions into cosine functions easily. So, pick \(u=\sin(x)\) and try to convert any remaining cosines using \(\cos^2(x)=1-\sin^2(x)\text{.}\)
You get a choice here! Both exponents are odd, so picking either function as \(u\) will leave you with an even exponent on the other function on which to use the Pythagorean Identity. Is there a choice of \(u\) that will be easier than the other choice?
Before you start thinking about this integral, let’s build the relevant version of the Pythagorean Identity that we’ll use. Our standard version of this is:
Since we want a version that connects \(\tan(x)\text{,}\) which also is written as \(\frac{\sin(x)}{\cos(x)}\text{,}\) with \(\sec(x)\text{,}\) or \(\frac{1}{\cos(x)}\text{,}\) let’s divide everything in the Pythagorean Identity by \(\cos^2(x)\text{:}\)
Now start with the integral. We’re going to use two different processes here, two different \(u\)-substitutions. First, set \(u=\tan(x)\text{.}\) Complete the substitution and solve the integral.
For each of these integrals, why were the exponents set up just right for \(u\)-substitution each time? How does the structure of the derivatives of each function play into this?
For integrals in the form\(\displaystyle\int\sec^p(x)\tan^q(x)\;dx\) where \(p\) and \(q\) are real number exponents:
If \(q\text{,}\) the exponent on \(\tan(x)\text{,}\) is odd, we can use \(u=\sec(x)\) and \(du=\sec(x)\tan(x)\;dx\text{.}\) Then we can apply the Pythagorean Identity \(\tan^2(x)=\sec^2(x)-1\text{.}\)
If \(p\text{,}\) the exponent on \(\sec(x)\text{,}\) is even, we can use \(u=\tan(x)\) and \(du=\sec^2(x)\;dx\text{.}\) Then we can apply the Pythagorean Identity \(\sec^2(x)=\tan^2(x)+1\text{.}\)
You have some choices here! If you use \(u=\sec(x)\text{,}\) then you’ll end up with a remaining \(tan^2(x)\) to convert using a Pythagorean Identity. Alternatively, if you use \(u=\tan(x)\text{,}\) you’ll end up needing to convert the remaining \(\sec^6(x) = (\sec^2(x))^3\text{.}\)
Another fraction exponent that is pushing us towards using \(u=\tan(x)\text{.}\) There will still be a \(\sec^4(x)\) to convert after substituting in \(du\text{.}\)
