Today’s quiz started with the indefinite integral \(\int\cos^2(t)\sin^5(t)\,dt\). First we must consider our options for u-substitution. We could go with \(u=\sin(t)\) or \(u=\cos(t)\) — but which one is best? Will they both work, maybe only one will… in these types of problems, everything is determined by our differential.
We start by considering \(u=\sin(t)\), with differential \(du=\cos(t)\,dt\). In giving up a factor of \(\cos(t)\) with \(dt\) to substitute for \(du\), we end up with \(\int\cos(t)u^5\,du\). We are left with only one factor of \(\cos(t)\), which we cannot exchange for \(sin(t)\) (using the pythagorean identities) because these exchanges must happen in pairs (i.e., we can exchange \(\cos^2(t)\), but not \(\cos(t)\)).
On the other hand, we may instead consider \(u=\cos(t)\), with differential \(du=-\sin(t)\,dt\) (which we re-write as \(-1\,du=\sin(t)\,dt\). In this case, we give up a factor of \(\sin(t)\) with \(dt\) to substitute for \(-1\,du\), ending up with \(\int u^2\sin^4(t)(-1)\,du=-1\int u^2\sin^4(t)\,du\). This situation is different than our first attempt, in that our remaining factors of \(\sin(t)\) are paired, and we can use the pythagorean identity to exchange \(\sin^2(t)\) for \(1-\cos^2(t)\). Since we have \(\sin^4(t)\), we can split it into pairs \(\sin^4(t)=\sin^2(t)\sin^2(t)\) and finally convert each pair: \[\sin^2(t)\sin^2(t)=(1-\cos^2(t))(1-\cos^2(t))=(1-u^2)(1-u^2)\]
So the full result of our substitution is \(-\int u^2(1-u^2)(1-u^2)\,du\). We cannot integrate products of functions, but in this case, we can perform the multiplication: \(u^2(1-u^2)(1-u^2)=u^2-2u^4+u^6\), which produces a function which we can integrate using the power rule: \(\frac13u^3-\frac25u^5+\frac17u^7+C\). Note that our integral has a coefficient of \(-1\), and also note that we must convert \(u\) back to \(\cos(t)\).
As a result, we have the final result: \[-\int u^2-2u^4+u^6\,du=-\frac13\cos^3(t)+\frac25\cos^5(t)-\frac17\cos^7(t)+C\]
In the second quiz problem, we need to find the antiderivative \(\int\tan^2(t)\sec^4(t)\,dt\). Again, we have two options, this time: \(u=\sec(t)\) or \(u=\tan(t)\). Recall that in the previous problem, the viability of each option was determined by what remained of our original function after accounting for the substitution of the differential \(du\). With this in mind, let’s look at what happens with the substitution of either differential.
If \(u=\sec(t)\), then \(du=\sec(t)\tan(t)\,dt\), and our integral looks like \(\int\tan(t)\sec^3(t)\,du\) after substituting only the differential. Since \(u=\sec(t)\), we are not too concerned about the odd power on \(\sec^3(t)\), since that would substitute to \(u^3\). However, the remaining \(\tan(t)\) is not good, since the pythagorean identities only allow us to exchange \(\tan^2(t)\), and not individual factors.
If instead, \(u=\tan(t)\), then \(du=\sec^2(t)\,dt\), and our integral looks like \(\int\tan^2(t)\sec^2(t)\,du\) after substituting only the differential. Since \(u=\tan(t)\), we know that \(\tan^2(t)\) will be substituted with \(u^2\). The remaining \(\sec^2(t)\) is good because we know that the pythagorean identity (tangent and secant form) says \(\sec^2(t)=\tan^2(t)+1\) which we can substitute with \(u^2+1\).
Since \(u=\tan(t)\) worked out for us, we now have the integral \[\int\tan^2(t)\sec^4(t)\,dt=\int u^2(u^2+1)\,du\]
As with our first problem, note that we cannot integrate this product as-is. Products must be dealt with before integrating — either with substitution, by parts, or (like in this case) simply the distributive property.
Since \(u^2(u^2+1)=u^4+u^2\), we can integrate the product: \[\int u^4+u^2\,du=\frac15u^5+\frac13u^3+C=\frac15\tan^5(t)+\frac13\tan^3(t)+C\]
Moving on to today’s topic, we look to work with integrals that contain factors such as \(\sqrt{N-x^2}\) (or, alternatively, \(\sqrt{x^2-N}\) or \(\sqrt{x^2+N}\)). Expressions such as these often lead us to attempt substitutions such as \(u=N-x^2\). However, if our integral cannot complete the substitution for the differential \(u=-2x\,dx\), we must have an alternative method. For example, \(\int\frac{1}{\sqrt{N-x^2}}\,dx\) does not have the necessary factor of \(x\,dx\) for substituting the differential.
Our goal then becomes to try and simplify \(\sqrt{N-x^2}\). This is the square-root of a difference (subtraction), which we cannot be assured of being a perfect square. For example \(\sqrt{25-x^2}\) cannot be reduced as-is, since we don’t know what is being subtracted from \(25\), and that difference might be a perfect-square, or it might not be. If \(x=3\), then that would be \(\sqrt{25-9}=\sqrt{16}=4\); but if \(x=1\), that would be \(\sqrt{25-1}=\sqrt{24}\) which does not simplify.
As mentioned (two paragraphs back), there are three radicals we seek to be able to simplify: \(\sqrt{N-x^2}\), \(\sqrt{x^2-N}\), and \(\sqrt{x^2+N}\) (which is equivalent to \(\sqrt{N+x^2}\)).
- If we have \(x^2+1\), we have a matching pattern in the pythagorean identity \(\tan^2(\theta)+1=\sec^2(\theta)\). Multiplying through by \(N\), we see that \(N\tan^2(\theta)+N=N\sec^2(\theta)\) — giving us an equivalence for our \(x^2+N\) form. Don’t worry about the differences between the two forms … we will deal with that shortly.
- If we instead have the form \(1-x^2\), we have the pythagorean identity \(1-\sin^2(\theta)=\cos^2(\theta)\). Multiplying through by \(N\), we see that \(N-N\sin^2(\theta)=N\cos^2(\theta)\) — giving an equivalence for our \(N-x^2\ form.
- Finally, if we have the form \(x^2-1\), we have the identity \(\sec^2(\theta)-1=\tan^2(\theta)\). Multiplying through by \(N\), we see that \(N\sec^2(\theta)-N=N\tan^2(\theta)\) — giving an equivalence for our \(x^2-N\) form.
Now let’s deal with the fact that we only have \(1-x^2\) when (for the pythagorean identity) we need something like \(1-\sin^2(\theta)\).
So far, when we think of substitution, we probably think of u-substitution, where we replace some function of \(x\) with a new variable, \(u\). In other words, \(u=f(x)\) allows us to swap out something complex (\(f(x)\)) for something simple (\(u\)). Today, we’re introducing another type of substitution, “Trig Substitution”, which replaces our variable \(x\) with a function of a new variable \(\theta\).
Specifically, in the case of \(1-x^2\), we want to replace \(x\) in such a way that we end up with \(1-\sin^2(\theta)\).
In order to accomplish this, we must make a substitution for \(x\) so that \(x^2\) is replaced with \(\sin^2(\theta)\). If you’re thinking this substitution is \(x=\sin(x)\), then you’re right!
Just like in the case of u-substitution, we cannot just swap over to a new variable — we must also make a substitution for the differential, \(dx\). In the case where we want to substitute \(x=\sin(\theta)\), the differential is \(dx=\cos(\theta)\,d\theta\).
[[Note the similarity in finding differentials for both forms of substitution]]
Consider an example such as \(\int\frac{1}{\sqrt{25-x^2}}\,dx\) where we observe the \(N-x^2\) form. Earlier, we noted how this form is related to the pythagorean identity \(1-\sin^2(\theta)=\cos^2(\theta)\).
Since we need (specifically) \(25-x^2\), that means we need to adjust our pythagorean identity by multiplying through by \(25\): \[25-25\sin^2(\theta)=25\cos^2(\theta)\]
Since our integral currently has \(25-x^2\), that means we need a substitution that will exchange our \(x^2\) with \(25\sin^2(\theta)\). What substitution should we make for \(x\)?
Since we need our \(x^2\) to be substituted with \(25\sin^2(\theta)\), our substitution needs to be \(x=5\sin(\theta)\). This substitution has differential \(dx=5\cos(\theta)\,d\theta\). When we substitute the integral: \[\int\frac{1}{\sqrt{25-x^2}}\,dx=\int\frac{1}{\sqrt{25-25\sin^2(\theta)}}(5\cos(\theta))\,d\theta\]
We can use the pythagorean identity to simplify \(25-25\sin^2(\theta)=25\cos^2(\theta)\).
We can further simplify \(\sqrt{25\cos^2(\theta)}=5\cos(\theta)\), and now our integral is \[\int\frac{1}{5\cos(\theta)}(5\cos(\theta))\,d\theta=\int 1\,d\theta=\theta+C\]
Finally, we must convert our antiderivative (in terms of \(\theta\)) into a function of our original variable, \(x\). Now, it’s not necessary in this particular example, but in general, we will need to know how to deal with this relationship between \(x\) and \(\theta\).
In this example, we had \(x=5\sin(\theta)\) — which we want to be able to express geometrically. \(\theta\) is an angle, and \(\sin(\theta)\) is then the sine ratio (opposite over hypotenuse) of sides in a right triangle with angle \(\theta\). Solving for \(sin(\theta)\), we get \(\sin(\theta)=\frac{x}{5}\).
We draw a right triangle (not to scale, because it doesn’t really matter), labelling one angle as \(\theta\), and identifying the opposite leg as \(x\) and the hypotenuse as \(5\). This leaves the adjacent leg as unkown, but we can use the pythagorean theorem: \(\text{adjacent}^2+x^2=5^2\), so the adjacent leg must be \(\sqrt{25-x^2}\).
This process is often necessary because our final antiderivative (in terms of \(\theta\)) might be something like \(\tan(\theta)\), which we would need to convert back into a function of \(x\). With a completed right triangle in hand, we can tell that \(tan(\theta)\) would be the opposite-over-adjacent ratio: \(\frac{x}{\sqrt{25-x^2}}\).
In our example, we need only to convert \(\theta+C\), so it is enough to solve \(\sin(\theta)=\frac{x}{5}\) for \(\theta\): \[\theta+C=\arcsin\left(\frac{x}{5}\right)+C\]
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