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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThanks for these tips.

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

Log in to reply

Same here xD

Log in to reply

Same pinch:)

Log in to reply

Dude thanks so much, this is great!

Log in to reply

This i an amazing post. Thank You ! :D

Log in to reply

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

Log in to reply

really helpful.

thanx for such note.

hope to see more notes from ya ;)

Log in to reply

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

Log in to reply

I'm not brainy enough to understand this!

Log in to reply