Our first quiz problem is a definite integral \(\displaystyle\int_3^5 2\sqrt{8x-15}\,dx\) and right off, we notice an easy linear substitution under a radical. So we aim to substitute \(u=8x-15\) and the corresponding differential \(du=8\,dx\). Now we don’t have a factor of \(8\) to go along with our \(dx\), but it’s just a constant, so there’s no issue restructuring the differential substitution to \(\frac18\,du=dx\).
After substituting, we’re left with \(\displaystyle\int_{x=3}^{x=5} 2\sqrt{u}\left(\frac18\,du\right)\) noting that our bounds are \(x\)-values (since the integral is now in terms of \(u\). We combine our coefficients to get \(\frac14\) times the integral of \(\sqrt{u}\). This anti-derivative uses the power rule, and we again combine coefficients: \(\frac14\) divided by \(\frac32\), which is \(\frac16\). Our antiderivative is then \(\frac16 u^\frac32=\frac16(8x-15)^\frac32\), which we then must evaluate at \(x=5\) and \(x=3\), then subtract the results: \[\frac16\left(25^\frac32-9^\frac32\right)=\frac16(125-27)=\frac{98}{6}=\frac{49}{3}\]
In the second quiz problem, we look at the indefinite integral \(\int\frac{(\ln 6x)^5}{x}\,dx\). There are two ways to approach this integral. In our first attempt, we identify \(u=6x\) as an “easy” substitution (it ends up being more work than expected, however). With \(u=6x\), we get differentials: \(du=6\,dx\), which we must manipulate to \(\frac16\,du=dx\).
The “problem” here is that while the argument of natural log is a straightforward replacement of \(6x\) with \(u\), the denominator is just \(x\) — not a \(6x\). This means our denominator must actually be replaced by \(\frac{u}{6}\). This means we end up with the integral \[\int\frac{(\ln u)^5}{\frac{u}{6}}\,\left(\frac16\,du\right)\]
We don’t like dividing by fractions (such as \(\frac{u}{6}\) here), so we instead multiply by the reciprocal \(\frac{6}{u}\) and simplify to end up with the integral \[\int\left(\ln u\right)^5\,\left(\frac{1}{u}\right)\,du\]
This is the integral of a product — which cannot be integrated in a straightforward manner. (Remember, the derivative of a product such as \(f(x)\cdot g(x)\) is NOT simply \(f'(x)\cdot g'(x)\).)
We must perform another substitution! Here we have \(\ln u\) in parenthesis (with another operation outside), so we aim for \(v=\ln u\) with \(dv = \frac{1}{u}\,du\).
After this substitution we finally end up with a straightforward power rule: \[\int v^5\,dv=\frac16 v^6+C=\frac16(\ln 6x)^6+C\]
Now, from the original integral \(\int\frac{(\ln 6x)^5}{x}\,dx\), we could have been more ambitious with our first substitution. By identifying \(u=\ln 6x\), we get \(du=\frac{1}{6x}\cdot 6\,dx\) (you do remember the chain rule, right?), which reduces to \(du=\frac{1}{x}\,dx\). After making this substitution, we immediately arrive at \[\int u^5\,du = \frac16 u^6+C = \frac16 \left(\ln 6x\right)^6+C\]
I hope this illustrates the potential benefits of making more ambitious (or ‘greedy’) substitutions.
In the third quiz problem, we are integrating \(\int x\cos(x^2)\sqrt{\sin(x^2)}\,dx\). Here we note that there is again more than one approach to substitution. If you immediately recognized \(u=x^2\), that’s a good start. However, that substitution is going to leave both sine and cosine to be dealt with — which will mean a second substitution. Instead, look deeper and find a more ambitious substitution.
What about substituting the radicand? \(u=\sin(x^2)\) is actually an excellent choice! It might not be immediately obvious until we look at the differential: \(du = \cos(x^2)\cdot 2x\,dx\). But why is this so good?
What factors do we need to be able to replace to end up with \(du\)? We need a factor of \(\cos(x^2)\), which we have. We also need a factor of \(2x\), which we almost have (remember that we can manipulate constant multiples easily) since we see \(x\) as a factor. And we also need \(dx\) (which we will always have).
With a tiny manipulation (dividing both sides by \(2\)), we get \(\frac12\,du=x\cos(x^2)\,dx\). With this substitution, we end up with a very straightforward integral: \(\frac12\int\sqrt{u}\,du\)
The antiderivative works out as follows: \[\frac12\int\sqrt{u}\,du=\frac12\frac23 u^\frac32+C=\frac13(\sin(x^2))^\frac32+C\]
It is very important to keep in mind that we cannot integrate products of functions. Think about what happens with derivatives (as antiderivatives tend to follow the same ‘rules’ about what ‘can’ or ‘cannot’ be done).
With derivatives, \(\frac{d}{dx}\left[f(x)\cdot g(x)\right]\neq f'(x)\cdot g'(x)\). This can be seen with a straightforward example. \(\frac{d}{dx}\left[x\cdot x\right] = \frac{d}{dx}\left[x^2\right]=2x\), but if we took the derivatives separately, \(\frac{d}{dx}[x]\cdot\frac{d}{dx}[x] = 1\cdot 1\). The two are clearly not equal.
Similarly, we cannot integrate a product of functions as the product of separate integrals.
So, how do we deal with integrals of products? Sometimes we can substitute, but that’s not always possible!
For derivatives, we find the derivative of the product of \(f(x)\) and \(g(x)\) using the product rule: \[\frac{d}{dx}\left[f(x)\cdot g(x)\right]=f'(x)\cdot g(x)+f(x)\cdot g'(x)\]
If we convert to the antiderivative form of this expression, we would say that \(f(x)\cdot g(x)\) is the antiderivative of \(f'(x)\cdot g(x)+f(x)\cdot g'(x)\). In mathematical notation this looks like \[\int\left(f'(x)\cdot g(x)+f(x)\cdot g'(x)\right)\,dx=f(x)\cdot g(x)\]
Now, we’d be pretty lucky to see an integral such as this… but in most cases our products are not so straightforward. We are frequently asked to integrate products that do not contain sums like \(\int (f’\cdot g+f\cdot g’)\,dx\) does.
Using the properties of integrals (similar to that of derivatives), we can rewrite \(\int (f’\cdot g+f\cdot g’)\,dx\) as \(\int f’\cdot g\,dx+\int f\cdot g’\,dx\), then solve our equation for \(\int f\cdot g’\,dx\):\[\int f\cdot g’\,dx = f\cdot g-\int f’\cdot g\,dx\]
It is important to note here that this equation only allows us to exchange the integral of one product (\(\int f\cdot g’\,dx\)) for the integral of a different product (\(\int g\cdot f’\,dx\)). This exchange technique is known as “Integration by Parts”.
Moving into an example of using Integration by Parts, we look at the integral \(\int x\sin(x)\,dx\) — identifying \(x\sin(x)\) as the product which we must separate into \(f(x)\) and \(g'(x)\).
It is important to recognize that the choice of which factor is \(f(x)\) and which is \(g'(x)\) is not an easy one when we are first learning this technique. At first, we might expect that \(f(x)=\sin(x)\) and \(g'(x)=x\,dx\). Since \(f'(x)\) is the derivative of \(f\), we get \(f'(x)=\cos(x)\,dx\); and \(g\) is the antiderivative of \(g’\), so \(g(x)=\frac12 x^2\). When we apply Integration by Parts: \(\int f\cdot g’ = f\cdot g – \int g\cdot f’\), so \[\int x\sin(x)\,dx = \frac12 x^2\sin(x)-\int\frac12 x^2\cos(x)\,dx\]
This new integral isn’t any better than the one we started with — it’s still a product of functions. This is a result of our choice for \(f\) and \(g’\). Perhaps we can do better with a different choice?
Choosing instead \(f(x)=x\) and \(g'(x)=\sin(x)\,dx\), we end up with \(f'(x)=dx\) and \(g(x)=-\cos(x)\). When we apply Integration by Parts to this alternative choice, we get \[\int x\sin(x)\,dx=-x\cos(x)-\int -\cos(x)\,dx\]
This new integral is no longer a product of functions (since \(f’\) is constant), meaning that we can evaluate \(\int g\cdot f’\)! \[-x\cos(x)-\int -\cos(x)\,dx=-x\cos(x)+\sin(x)+C\]
This antiderivative can be confirmed by taking the derivative and seeing that our result simplifies to be \(\frac{d}{dx}\left[-x\cos(x)+\sin(x)\right]=x\sin(x)\). Note the necessary use of the product rule in computing this derivative.
Finally, we look at another (odd) example. This example does not appear to be a product (as would usually trigger our attempts at Integration by Parts), but it is instead an integral of a “simple” function without a known antiderivative (the natural logarithm).
Since the antiderivative of \(\ln x\) is unknown, we need some other technique here. There’s nothing available to substitute, so we look to Integrating by Parts here. The question is, with no product, what two factors will we have? Simply put, one of the factors will be \(1\), and the other will be \(\ln x\).
Note that it is impossible for us to select \(\ln x\) for \(g’\) because we don’t know an antiderivative for \(\ln x\). That leaves us with the choice of \(f(x)=\ln x\) and \(g'(x)=1\,dx\). That means we end up with \(f'(x)=\frac{1}{x}\,dx\) and \(g(x)=x\).
Now we apply Integration by Parts and end up with \[\int\ln(x)\,dx=x\ln(x)-\int x\frac{1}{x}\,dx=x\ln(x)-\int 1\,dx=x\ln(x)-x+C\]
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