# 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.

Note by Trevor B.
4 years, 11 months ago

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Thanks for these tips.
I almost misread the title, thinking that it is a Computer Science post (get it?, because C++)

- 4 years, 11 months ago

Same here xD

- 4 years ago

Same pinch:)

- 3 years, 6 months ago

Dude thanks so much, this is great!

- 4 years, 11 months ago

This i an amazing post. Thank You ! :D

- 4 years, 11 months ago

Aye, very important to realise that we lose information when taking derivatives. Great post!

- 4 years, 11 months ago

thanx for such note.

hope to see more notes from ya ;)

- 4 years ago

In the equation below domain of C, $$3x^2-2x$$ one! You have an extra "=" sign which was confusing me!

- 4 years ago

I'm not brainy enough to understand this!

- 3 years, 6 months ago

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