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Let us consider the following integral $$\int \frac{\sin x}{\cos^3 x}dx\, .$$ If we note that $\tan x= \frac{\sin x}{\cos x}$, we have $$\int \frac{\tan x}{\cos^2 x}dx = \frac{\tan^2 x}{2}+C \quad (1)$$ since $\int f(x) f^\prime (x) dx= \frac{f^2(x)}{2}+C$ and $f(x)=\tan x$.

Otherwise, if we substitute $t=\cos x$: $$-\int\frac{dt}{t^3}=\frac{1}{2\cos^2 x}+C \quad (2) .$$ The result shall be the same, but $\frac{\tan^2 x}{2}\neq \frac{1}{2\cos^2 x}$. The reason for this apparent difference is the constant $C$ in (1) and (2). They are not the same constant, and actually $\frac{1}{\cos^2 x}=\tan^2 x+1$. If we denote by $C_1$ the constant of (1) and by $C_2$ the constant of (2), we have $C_2=\frac12+C_1$. In my opinion, this example is useful to let students know how important is the constant of integration.

Can someone suggest me other examples of this type, without trigonometric functions?

Mark
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    This is probably the most well-known example, and certain classes of functions (e.g. polynomials) couldn't lead to such examples because of how we antidifferentiate them. On a separate note, I would recommend changing your title to something much more specific like "Is there an non-trig example of two antiderivatives calculated by hand that are not equal?" – Mark S. Nov 08 '18 at 15:42
  • Some thoughts. In either case, (1) contains all the antiderivatives and (2) contains all antiderivatives. $\frac{x^3}{3}+5$ and $\frac{x^3}{3}+7$ are specific antiderivatives of $x^2$. Each specific antiderivative can be made into a general antiderivative as $\frac{x^3}{3}+5 +C$ and $\frac{x^3}{3}+ 7 + C$. As written, I never intended to equate both general antiderivatives because then 5 = 7. If I want to equate two general antiderivatives (whose specific antiderivatives are different), either both need to have different letters for the constant (give one C and the other D) or add in another – DWade64 Nov 08 '18 at 17:25
  • constant like $\frac{x^3}{3}+ 5 + C = \frac{x^3}{3}+7+C+D$. In your specific example, your constants tell you a shift, referenced from/based on the specific antiderivative shown. In $\frac{x^3}{3}+5 +C$, $C$ shows a shift from the specific antiderivative $\frac{x^3}{3}+ 5$. A $C$ of 2 is for a shift up of the specific antiderivative $\frac{x^3}{3}+5$. If my specific antiderivative is different, then if I want my graphs to be the same, the constant will have to be different – DWade64 Nov 08 '18 at 17:27
  • The only reason to equate general antiderivatives is to find the constant such that you are always talking about the same specific antiderivative at all times. In your example, you found that given $C_1$, $C_2$ would have to be 1/2 + $C_1$ so that you are talking about the same antiderivative at all times (but which antiderivative you are talking about is never important for the Theorem of Calculus) – DWade64 Nov 08 '18 at 17:38

2 Answers2

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You can also refer to these other interesting cases

user
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One interesting thing is to take a look at the expansion of $e^x$: $$e^x=\sum_{k\ge0}\frac{x^k}{k!}$$If we take the integral of both sides, we get $$\sum_{k\ge1}\frac{x^k}{k!}+C$$Notice the change in the index. This is because the number $1$ becomes $x$ after integration. But because of the constant of integration, $e^x$ is the antiderivative of itself.

Kamal Saleh
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