Integral Rules¶

Every derivative rule tells something about antiderivatives. Next we explore what we can do with different derivative rules that we know.

Antiderivatives of sin and cos¶

Because the derivative of $\sin(x)$ is $\cos(x)$, we get the following result.

On the other hand, $\frac{d}{dx}\cos(x) = -\sin(x)$, with an annoying minus sign. To get rid of that, we can add another minus sign in front of $\cos(x)$, and the minuses will cancel: $$ \begin{align} \frac{d}{dx}(-\cos(x)) &= (-1)\frac{d}{dx}\cos(x)\\ &= (-1)(-\sin(x)) = \sin(x) \end{align} $$

I don't recommend memorizing these integral rules; it's better to learn deriving them from the derivative rules in your head. We will find the antiderivative of $\tan$ later.

Sum¶

Let's say you want to find the antiderivatives of $2x + \cos(x)$. Because $$ \frac{d}{dx}(x^2 + \sin(x)) = \frac{d}{dx} x^2 + \frac{d}{dx} \sin(x) = 2x+\cos(x), $$ we get $$ \int(2x+\cos(x))\ dx = x^2+\sin(x)+C. $$ More generally, if $F$ is an antiderivative of $f$ and $G$ is an antiderivative of $g$, then $$ \int (f(x)+g(x))\ dx = F(x)+G(x)+C. $$ Usually this result is written like this:

Each $\int \dots \ dx$ on the right side creates its own $+C$, so technically using this rule should look like this: $$ \begin{align} \int (2x+\cos(x))\ dx &= \int 2x\ dx + \int \cos(x)\ dx \\ &= (x^2+C_1)+(\sin(x)+C_2) \\ &= x^2+\sin(x)+\underbrace{(C_1+C_2)}_{\text{let's call this $C$}} \\ &= x^2+\sin(x)+C \end{align} $$ Here $C_1$ came from $\int 2x\ dx$ and $C_2$ came from $\int \cos(x)\ dx$. Letting $C=C_1+C_2$ simplified the result. In practice, that is unnecessary, and we just write $$ \int (2x+\cos(x))\ dx = \int 2x\ dx + \int \cos(x)\ dx = x^2+\sin(x)+C. $$

Let's see if the same rule works for integrals (as in area, not integrals as in antiderivative). For using part 2 of the fundamental theorem of calculus, we assume that $f$ and $g$ are continuous and they have antiderivatives $F$ and $G$. Then, with handy notation, we get $$ \begin{align} \int_a^b (f(x)+g(x))\ dx &= \left[ F(x)+G(x) \right]_a^b \\ &= (F(b)+G(b))-(F(a)+G(a)) \\ &= F(b)-F(a)+G(b)-G(a) \\ &= \int_a^bf(x)\ dx + \int_a^bg(x)\ dx. \end{align} $$

Constant in front¶

For any constant $k$ and function $f$ with antiderivative $F$, we have $$ \frac{d}{dx} (kF(x)) = kF'(x) = kf(x), $$ so $$ \int kf(x)\ dx = kF(x)+C. $$ This result is typically written like this:

Just like with derivatives, this does not work if $k$ is not a constant. For example, $$ \int 3x \cdot 2x\ dx = \int 6x^2\ dx = 2x^3 + C, $$ but $$ 3x\int 2x\ dx = 3x(x^2 + C) = 3x^3 + 3Cx. $$

Also, the above way to write this result is slightly broken. If $k=0$, then the left side is $$ \int 0f(x)\ dx = \int 0\ dx = C, $$ but the right side is $$ 0\int f(x)\ dx = 0, $$ because multiplying anything with $0$ always gives $0$. This can be solved by simply writing one $+C$ at the end of the antiderivative calculation, e.g. like this: $$ \int 0\cos(x)\ dx = 0\int \cos(x)\ dx = 0\sin(x) + C = C $$

For integrals (not antiderivatives), we get $$ \begin{align} \int_a^b kf(x)\ dx &= \left[ kF(x) \right]_a^b \\ &= kF(b)-kF(a) \\ &= k(F(b)-F(a)) \\ &= k\int_a^b f(x)\ dx, \end{align} $$ and in this calculation, nothing breaks if $k=0$.

Antiderivatives of $x^n$ for $n \ne -1$¶

Let's calculate the antiderivative of $x^5$. According to the power rule $$ \frac{d}{dx} x^6 = 6x^5. $$ To get rid of $6$, we can multiply by $\frac{1}{6}$ to cancel it, and we get $$ \frac{d}{dx}\left( \frac{1}{6}x^6 \right) = \frac{1}{6} 6x^5 = x^5. $$ With any other number $n$ instead of $5$, we have $n+1$ instead of $6$. To avoid division by zero in $\frac{1}{n+1}$, we need to assume that $n \ne -1$.

Antiderivatives of $x^{-1}$¶

The above calculation does not work for $n=-1$. This does not mean that the function $f(x)=x^{-1}=1/x$ does not have an antiderivative. In fact, that function is continuous everywhere except at zero, so it has some antiderivative everywhere except at zero.

From logarithms (TODO), we know that $$ \frac{d}{dx} \ln(x) = \frac{1}{x}, $$ so $$ \int \frac{1}{x}\ dx = \ln(x). $$ However, this only works when $x$ is positive, because $\ln(x)$ is not defined for $x \le 0$. If $x$ is negative, then $-x=\abs{x}$ is positive, and with the chain rule, we get $$ \frac{d}{dx} \ln(-x) = \frac{1}{-x} \frac{d}{dx}(-x) = \frac{1}{-x}(-1) = \frac{1}{x}. $$ So, we have $$ \int \frac{1}{x}\ dx = \begin{cases} \ln(x)+C, & \text{if $x > 0$} \\ \ln(-x)+C, & \text{if $x < 0$.} \end{cases} $$ These cases can be combined together with absolute value, like this:

You need to be somewhat careful when using the above result. For example, it might be tempting to do this: $$ \int_{-2}^2 \frac{1}{x}\ dx = \left[ \ln\abs{x} \right]_{-2}^2 = \ln\abs{2} - \ln\abs{-2} = \ln(2)-\ln(2) = 0 $$ This is wrong. Because $\frac{1}{x}$ is undefined at $0$, it's not integrable on $[-2,2]$, and the fundamental theorem of calculus doesn't apply. In general, we have $$ \int_a^b \frac{1}{x}\ dx = \begin{cases} \ln\abs{b}-\ln\abs{a}, & \text{if $a$ and $b$ are both positive or both negative} \\ \text{undefined} & \text{in all other cases}. \end{cases} $$

Antiderivatives of exponent functions¶

Because $\frac{d}{dx} e^x = e^x$ (TODO), we get the following result.

More generally, for any positive base $b$, the derivative of $b^x$ is $b^x \ln(b)$ (TODO). So, if $\ln(b) \ne 0$, we have $$ \frac{d}{dx} \left( \frac{1}{\ln(b)}b^x \right) = \frac{1}{\ln(b)} \frac{d}{dx} b^x = \frac{1}{\ln(b)}b^x\ln(b) = b^x. $$ Note that $\ln(1) = 0$. Because $\ln$ is strictly increasing, this is the only way how $\ln$ can be zero.

For $b=1$, we have $b^x = 1^x = 1$ and $\int 1\ dx = x+C$. Also note that $\ln(e)=\ln(e^1)=1$, so for $b=e$, the two results give the same formula.