# Calculus

Antiderivatives

Integral calculus can be divided into two types of integrals: the definite and the indefinite. The indefinite integral is also called the antiderivative. If you can master the antiderivative, then definite integrals will be cake for you.

Intuitively, if you think of the derivative of a function as making the function more and more simple, the antiderivative is just the opposite; it makes the function more complex (i.e. if for the derivative of a polynomial you move down a power from the parent function, if you were to integrate the same function, you would move up a power instead).

For our purposes, the antiderivative is the algebraic calculation of the function of an integral. Basically, an antiderivative is a function, $F(x)$, whose derivative is $f(x)$ (that is, $F'(x)=f(x)$).

Symbols and Terms

Integrate means “find the integral”
$\displaystyle\int$ is the integration symbol (operator)
Like in derivatives, $dx$ means "with respect to x"

A properly set-up integral looks like this: $\int{f(x)}dx=F(x)+C$. We refer to $f(x)$ as the integrand, meaning "that which is being integrated". $C$ is the constant of integration. It is necessary to always add an arbitrary constant to the end of your integral, as this will always drop when taking the derivative of a function. If you do not add the $+C$ you are assuming that the constant added to the function is zero, which is rarely the case.

Do Not Forget Your $+C$

Common Integrals

Note that you can use your knowledge of trigonometric derivatives to understand integrals (i.e. $\displaystyle\int \cos(x)dx = \sin(x)+C, \int \sec^2(x)dx=\tan(x)+C$, etc.). Here are some more integrals that you should know:
\begin{array}{|c|c|} \hline f(x) & F(x) \\ \hline x^n & \frac{x^{n+1}}{n+1}+C\\ \hline e^x & e^x+C\\ \hline \frac{1}{x} & \ln|x|+C\\ \hline \ln(x) & x\ln(x)-x+C\\ \hline a^x & \frac{a^x}{\ln a}+C\\ \hline \sin(x) & -\cos(x)+C\\ \hline \cos(x) & -\sin(x)+C\\ \hline \tan(x) & \ln |\sec(x)|+C\\ {} & -\ln |\cos(x)|+C\\ \hline \cot(x) & \ln|\sin(x)|+C\\ {} & -\ln|\csc(x)|+C\\ \hline \sec(x) & \ln|\sec(x)+\tan(x)|+C\\ {} & -\ln|\sec(x)-\tan(x)|+C\\ \hline \csc(x) & \ln|\csc(x)-\cot(x)|+C\\ {} & -\ln |\csc(x)+\cot(x)|+C\\ \hline \end{array}

Properties of Indefinite Integrals

The main properties of indefinite integrals to remember are:

A) $\displaystyle\int f(x) \pm g(x)dx= \int f(x)dx \pm \int g(x)dx$
B) $\displaystyle\int k \cdot f(x)dx = k\int f(x)dx$
C) $\displaystyle\int k dx = kx+C$

Techniques of Integration

Reverse Power Rule
We know that for derivatives, $\frac{d}{dx} x^n=nx^{n-1}$. For integrals, the formula is $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Example: $\displaystyle\int x^3 dx = \frac{x^4}{4}+C$

U-Substitution
U-Substitution is as important to integration as the chain rule is to differentiation; in essence, U-Substitution "unravels" the chain rule.

When to use it: U-Substitution is used when the integrand contains multiple functions that are related by differentiation. Generally, if you see an integral whose integrand is comprised of the product of a composite and non-composite functions, and if one of the functions in the integrand looks as if it could be the derivative of another function in the integrand. A trick for choosing $u$ is to look for a function which involves addition or subtraction and making that $=u$, effectively getting rid of sloppy sums.

How to use it: This is a long paragraph but make sure to read it all. I'll use the example of $\displaystyle\int 2x\cos(x^2)dx$. First, locate the two functions which share a function-derivative relationship. By looking at our function we can see that these two functions will be $x^2$ and $2x$. Next, use the letter "u" to represent the parent function, that is, set $u=x^2$. This implies that $\frac{du}{dx}=2x$. Write these down - it only takes a second and it will help you remember them later. Now, substitute this into your original integral to get $\displaystyle\int \frac{du}{dx} \cos(u)dx=\displaystyle\int \cos(u) \frac{du}{dx} dx = \displaystyle\int cos(u) du$. Although it is not really a fraction, $\frac{du}{dx}$ can be treated like one in this case; thuse multiplying it by $dx$ will cancel out the denominator and leave just $du$. This will always be the case. Now all we are left with is simply the integral of a cosine function, which we know to be the sine function. Thus, $\displaystyle\int \cos(u)du= \sin(u)+C$ (if you are confused, see Common Integrals). All that's left now is substituting back: $\sin(u)+C=\sin (x^2)+C$ since $u=x^2$ as defined before.

Oftentimes, the function-derivative relationship may not be so obvious, and you may need to do extra work to get the integral in terms of $u$. One way of doing this is by pulling a constant out of the integral. Here is an example.

Find $\displaystyle\int x\sec^2(x^2)dx$. Setting $u=x^2$, we get $du=2x$. However, instead of $2x$, we only have $x$ in the integrand. Luckily, if we are crafty, we can see that we can rewrite this integral as $\displaystyle\int \sec^2(u)\frac{1}{2}du=\displaystyle\int \frac{1}{2}\sec^2(u)du$. To make integrals easier, we always pull out as many constants as we can, so this expression becomes $\frac{1}{2}\displaystyle\int \sec^2(u)du$, which is simply equal to $\frac{1}{2}(\tan(u)+C=\frac{1}{2}\tan(x^2)+C$. Wasn't that fun?

U-substitution can be done more than once in the same integral (in this case, you might want to change the letter to w)

If you split an integral with a sum into two integrals, you can use different $u$s for both

If, say, you have $u=x-2$, then this implies that $x=u-2$. This is called “back substitution" and can be of real use sometimes. Simply substitute the $x$-equivalent to get the function in terms of $u$ and solve.

Remember: don't be afraid to get messy and rearrange the integral; oftentimes, the most simplified form is not the easiest to work with