The Importance of +C

We've all done it. Anyone who's taken calculus in their life has at some point forgotten to put the $+C$ at the end of an indefinite integral. But there's more to the $+C$ then just something you have to remember or you'll lose points on your calculus test. I'm going to discuss a couple integrals and explain how the $+C$ can help you avoid mathematical disaster. $\text{............................................................................................}$ First, an explanation of the $+C.$ What is its purpose? When you take an indefinite integral of $f(x)$ you are figuring out what function $F(x)$ has the derivative equal to $f(x).$ Take a look at two functions, $f_1(x)=x^3$ and $f_2(x)=x^3+1.$ Obviously, these are different functions, but when you derive them, you will find that $f_1'(x)=f_2'(x)=3x^2!$ So if we are trying to find the antiderivative of $3x^2,$ we add a $+C,$ meaning adding a constant to the antiderivative, to account for the fact that $f(x)+C$ derives to $f'(x)$ for all real $C.$ $\text{............................................................................................}$ So let's take a look at a couple integrals.

This first one has several different ways to find the antiderivative. $\int\sin x\cos x\text{ }dx$ Let's start with my preferred way. $\int\sin x\cos x\text{ }dx=\int\dfrac{1}{2}\sin2x\text{ }dx=-\dfrac{1}{4}\cos 2x+C\text{ }\text{ }(1)$ But we can make a couple different $u\text{-substitutions}$ too. Let $u=\sin x$ and $du=\cos x\text{ }dx.$ $\int\sin x\cos x\text{ }dx=\int u\text{ }du=\dfrac{1}{2}u^2+C=\dfrac{1}{2}\sin^2x+C\text{ }\text{ }(2)$ Now let $u=\cos x$ and $du=-\sin x\text{ }dx.$ $\int\sin x\cos x\text{ }dx=\int-u\text{ }du=-\dfrac{1}{2}u^2+C=-\dfrac{1}{2}\cos^2x+C\text{ }\text{ }(3)$ Somehow, these are all equal to each other. But how? Let's take a look at some properties of $C$ before coming back to these integrals. $\text{............................................................................................}$ Remember that $C$ is a constant between $-\infty$ and $\infty.$ Don't look at $C$ as a number, but a function of with a range of $(-\infty,\infty).$ So we can say, for example, this. $\int\sin x\cos x\text{ }dx=1-\dfrac{1}{2}\cos^2x+C$ $\textit{and}\int\sin x\cos x\text{ }dx=-\dfrac{1}{2}\cos^2x+C-C^3+\sqrt[1337]{C}$ As long as the function of $C$ has range $(-\infty,\infty),$ then you can use it, but you may need to restrict the domain. Take a look at this. $\int3x^2\text{ }dx=x^3+\tan C\text{ if }C\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ One last thing to note before looking at the integral again is that not all $C\text{'s}$ are the same. Subtraction of these integrals does not take away the $C.$ Here is the misconception. $\int3x^2\text{ }dx-\int2x\text{ }dx=x^3+C-(x^2+C)=x^3-x^2$ But those $C\text{'s}$ are $\textit{not the same}.$ A constant minus a constant is another constant. $\int3x^2\text{ }dx-\int2x\text{ }dx=x^3+C_1-(x^2+C_2)=x^3-x^2+C_3$ This makes sense because of this. $\int3x^2-2x\text{ }dx=x^3-x^2+C$ So we've found some properties of the constants in integrals. Let's go back to the integral of $\sin x\cos x.$ $\text{............................................................................................}$ We have already proved that $(2)=(3).$ Let's find the constant that will make $(1)=(2)$

$-\frac{1}{4}\cos 2x+C_1=\frac{1}{2}\sin^2x+C_2\Rightarrow-\frac{1}{4}\cos 2x+C_3=\frac{1}{2}\sin^2x$

Expanding the LHS, $-\frac{1}{4}+\frac{1}{2}\sin^2x+C_3=\frac{1}{2}\sin^2x$

Look at that! Cancel out the $\frac{1}{2}\sin^2x$ to find that $-\frac{1}{4}+C_3=0,$ so $C_3=\frac{1}{4}.$ The difference between the constants of the two integrals is $\frac{1}{4}.$

So let's conclude our proof that $(1)=(2)$ by saying this. Here, the $C\text{'s}$ are equal to each other. $\int\sin x\cos x\text{ }dx=\int\sin \cos x\text{ }dx$ $-\dfrac{1}{4}\cos 2x+\frac{1}{4}+C=\dfrac{1}{2}\sin^2x+C$ $\text{............................................................................................}$ Not all integrals in need of recalibrating the $+C$ involve manipulating multiple $u\text{-substitutions.}$ The next one we will work with is this. $\int\dfrac{e^x+e^{-x}}{e^x-e^{-x}}\text{ }dx$ Let $u=e^x-e^{-x}.$ Then $du=(e^x+e^{-x})dx$ $\int\dfrac{du}{u}=\ln|u|+C_1=\ln(e^x-e^{-x})+C_1$ But that integral looks like it uses hyperbolic geometry. In fact, it is simply this. $\int\coth x\text{ }dx=\ln(\sinh x)+C_2=\ln\left(\dfrac{e^x-e^{-x}}{2}\right)+C_2$ We have another case of needing to manipulate the $+C\text{'s}.$

Using the properties of logarithms, $\ln\left(\frac{e^x-e^{-x}}{2}\right)=\ln(e^x-e^{-x})-\ln2.$ But $\ln2$ is a constant. So $C_1=C_2+\ln2$ and the integrals are equal! $\int\dfrac{e^x+e^{-x}}{e^x-e^{-x}}\text{ }dx=\int\dfrac{e^x+e^{-x}}{e^x-e^{-x}}\text{ }dx$ $\ln\left(\frac{e^x-e^{-x}}{2}\right)+\ln2+C=\ln(e^x-e^{-x})+C$ $\text{............................................................................................}$ So now we've seen how we can manipulate the $+C$ in an indefinite integral to make sure that the laws of math aren't broken. Here are some strategies for manipulating the constant.

$-$Separate the $C\text{'s}$ with subscripts. Not all constants are the same.

$-$The difference between two constants is still a constant that may or may not be $0.$

$-$$C$ represents a function with range $\mathbb{R}$

$-$Often, you can use properties of logarithms (or other functions) to separate out constants from a variable expression.

$-$If the simplified version of your indefinite integral contains an integer added to a variable expression, you may want to add the integer to $C$ to produce a new $C.$ $\text{............................................................................................}$ $\textbf{Problem:}$ Discuss this integral with respect to manipulating the constants. $\int x^2\sin(x^3)\cos(x^3)\text{ }dx$ Can you share a problem that uses this type of variable manipulation? Do you have any comments about this topic? Please comment and leave your ideas. $\text{............................................................................................}$ Thanks for reading this post, and I hope it helps out! I'll be using the hashtag #TrevorsTips for posts that talk about seemingly simple concepts that can actually be really helpful.