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.

## Comments

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TopNewestThanks for these tips.

I almost misread the title, thinking that it is a Computer Science post (get it?, because C++) – Anish Puthuraya · 3 years, 5 months ago

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– Ashley Shamidha · 2 years ago

Same pinch:)Log in to reply

– Keshav Tiwari · 2 years, 7 months ago

Same here xDLog in to reply

I'm not brainy enough to understand this! – Ashley Shamidha · 2 years ago

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In the equation below domain of C, \(3x^2-2x\) one! You have an extra "=" sign which was confusing me! – Pranjal Jain · 2 years, 6 months ago

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really helpful.

thanx for such note.

hope to see more notes from ya ;) – Soumo Mukherjee · 2 years, 7 months ago

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Aye, very important to realise that we lose information when taking derivatives. Great post! – Vishnuram Leonardodavinci · 3 years, 5 months ago

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This i an amazing post. Thank You ! :D – Priyansh Sangule · 3 years, 5 months ago

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Dude thanks so much, this is great! – Finn Hulse · 3 years, 5 months ago

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