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+C at the end of an indefinite integral. But there's more to the +C+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+C can help you avoid mathematical disaster. ............................................................................................\text{............................................................................................} First, an explanation of the +C.+C. What is its purpose? When you take an indefinite integral of f(x)f(x) you are figuring out what function F(x)F(x) has the derivative equal to f(x).f(x). Take a look at two functions, f1(x)=x3f_1(x)=x^3 and f2(x)=x3+1.f_2(x)=x^3+1. Obviously, these are different functions, but when you derive them, you will find that f1(x)=f2(x)=3x2!f_1'(x)=f_2'(x)=3x^2! So if we are trying to find the antiderivative of 3x2,3x^2, we add a +C,+C, meaning adding a constant to the antiderivative, to account for the fact that f(x)+Cf(x)+C derives to f(x)f'(x) for all real C.C. ............................................................................................\text{............................................................................................} So let's take a look at a couple integrals.

This first one has several different ways to find the antiderivative. sinxcosx dx\int\sin x\cos x\text{ }dx Let's start with my preferred way. sinxcosx dx=12sin2x dx=14cos2x+C  (1)\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-substitutionsu\text{-substitutions} too. Let u=sinxu=\sin x and du=cosx dx.du=\cos x\text{ }dx. sinxcosx dx=u du=12u2+C=12sin2x+C  (2)\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=cosxu=\cos x and du=sinx dx.du=-\sin x\text{ }dx. sinxcosx dx=u du=12u2+C=12cos2x+C  (3)\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 CC before coming back to these integrals. ............................................................................................\text{............................................................................................} Remember that CC is a constant between -\infty and .\infty. Don't look at CC as a number, but a function of with a range of (,).(-\infty,\infty). So we can say, for example, this. sinxcosx dx=112cos2x+C\int\sin x\cos x\text{ }dx=1-\dfrac{1}{2}\cos^2x+C andsinxcosx dx=12cos2x+CC3+C1337\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 CC has range (,),(-\infty,\infty), then you can use it, but you may need to restrict the domain. Take a look at this. 3x2 dx=x3+tanC if C(π2,π2)\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’sC\text{'s} are the same. Subtraction of these integrals does not take away the C.C. Here is the misconception. 3x2 dx2x dx=x3+C(x2+C)=x3x2\int3x^2\text{ }dx-\int2x\text{ }dx=x^3+C-(x^2+C)=x^3-x^2 But those C’sC\text{'s} are not the same.\textit{not the same}. A constant minus a constant is another constant. 3x2 dx2x dx=x3+C1(x2+C2)=x3x2+C3\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. 3x22x dx=x3x2+C\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 sinxcosx.\sin x\cos x. ............................................................................................\text{............................................................................................} We have already proved that (2)=(3).(2)=(3). Let's find the constant that will make (1)=(2)(1)=(2)

14cos2x+C1=12sin2x+C214cos2x+C3=12sin2x-\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, 14+12sin2x+C3=12sin2x-\frac{1}{4}+\frac{1}{2}\sin^2x+C_3=\frac{1}{2}\sin^2x

Look at that! Cancel out the 12sin2x\frac{1}{2}\sin^2x to find that 14+C3=0,-\frac{1}{4}+C_3=0, so C3=14.C_3=\frac{1}{4}. The difference between the constants of the two integrals is 14.\frac{1}{4}.

So let's conclude our proof that (1)=(2)(1)=(2) by saying this. Here, the C’sC\text{'s} are equal to each other. sinxcosx dx=sincosx dx\int\sin x\cos x\text{ }dx=\int\sin \cos x\text{ }dx 14cos2x+14+C=12sin2x+C-\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+C involve manipulating multiple u-substitutions.u\text{-substitutions.} The next one we will work with is this. ex+exexex dx\int\dfrac{e^x+e^{-x}}{e^x-e^{-x}}\text{ }dx Let u=exex.u=e^x-e^{-x}. Then du=(ex+ex)dxdu=(e^x+e^{-x})dx duu=lnu+C1=ln(exex)+C1\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. cothx dx=ln(sinhx)+C2=ln(exex2)+C2\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’s.+C\text{'s}.

Using the properties of logarithms, ln(exex2)=ln(exex)ln2.\ln\left(\frac{e^x-e^{-x}}{2}\right)=\ln(e^x-e^{-x})-\ln2. But ln2\ln2 is a constant. So C1=C2+ln2C_1=C_2+\ln2 and the integrals are equal! ex+exexex dx=ex+exexex dx\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(exex2)+ln2+C=ln(exex)+C\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+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’sC\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 0.0.

-CC represents a function with range R\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 CC to produce a new C.C. ............................................................................................\text{............................................................................................} Problem:\textbf{Problem:} Discuss this integral with respect to manipulating the constants. x2sin(x3)cos(x3) dx\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.
5 years, 8 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++)

Anish Puthuraya - 5 years, 8 months ago

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Same here xD

Keshav Tiwari - 4 years, 9 months ago

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Same pinch:)

Ashley Shamidha - 4 years, 3 months ago

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Dude thanks so much, this is great!

Finn Hulse - 5 years, 8 months ago

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This i an amazing post. Thank You ! :D

Priyansh Sangule - 5 years, 8 months ago

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

Vishnuram Leonardodavinci - 5 years, 8 months ago

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really helpful.

thanx for such note.

hope to see more notes from ya ;)

Soumo Mukherjee - 4 years, 9 months ago

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In the equation below domain of C, 3x22x3x^2-2x one! You have an extra "=" sign which was confusing me!

Pranjal Jain - 4 years, 9 months ago

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I'm not brainy enough to understand this!

Ashley Shamidha - 4 years, 3 months ago

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