Derivatives find instantaneous rates of change. By contrast, integrals find areas under curves.

The integral symbol $\int$ is derived from the word “sum.” To estimate the area under the curve, we can approximate the curve using rectangles and compute the sum of their area, as pictured below.

This is called a Riemann sum. Dr. John Urschel explains Riemann sum in this lesson clip from his Outlier course:

However, this sum either overestimates or underestimates the area under the curve.

To compute the precise area under the curve, we take the limit of the Riemann sum as the number of subdivisions approaches infinity. We can think of this limit as the sum of infinitely many rectangles, where each rectangle’s height is $y$, and each rectangle’s width is a tiny change in $x$. This gives us the definition of a definite integral, where $A$ represents the area under the curve on $[a, b]$.

$\int_{a}^{b} f(x)\,dx = A$

What does $\int$ mean? The symbol $\int$ is called the integral sign. The inside function $f(x)$ is called the integrand.

To evaluate a definite integral, we take the difference between the indefinite integral of the function evaluated at $a$ and the indefinite integral of the function evaluated at $b$.

An indefinite integral finds the antiderivatives of a function, which we usually denote by $F(x)$:

$\int f(x)\,dx = F(x) + C$

If $F’(x) = f(x)$, then $F(x)$ is an antiderivative of $f(x)$. This means that taking the derivative of $F(x)$ gives back $f(x)$. Evaluating an indefinite integral answers the question, “What functions, when differentiated, give back the integrand we started with?”

You can think about this process as differentiation in reverse. If $\int f(x)\,dx = F(x) + C$, then $F’(x) = f(x)$. As mentioned earlier, the antiderivative function $F(x)$ can be used to calculate all possible areas under the curve by taking the difference of $F(x)$ evaluated at the endpoints of an interval.

It might be helpful to think about this concept another way, too. If integrating a derivative of a function, then we have:

$\int f’(x)\,dx = f(x) + C$

$C$ represents the constant of integration. Since the derivative of any constant is zero, $C$ can take any value. Thus, the indefinite integral gives us a family of antiderivative functions.

Top 6 Integral Shortcuts

How do you integrate quickly? We’ll cover 6 handy tricks for evaluating integrals.

1. Memorize Basic Integral Formulas

The most practical integral trick is to memorize the most common integral examples. Over time, these examples have become standard integral equations. These basic integral formulas are essential tools for evaluating integrals.

5 Basic Integration Formulas

Here are some basic integration formulas. These equations are given below.

$\int \sin{(ax)}\,dx = \frac{-\cos{(ax)}}{a}+C$ for some real number $a$

$\int \cos{(ax)}\,dx = \frac{\sin{(ax)}}{a}+C$ for some real number $a$

Absolute Value Rule

$\int |x|\,dx = \frac{x |x|}{2} + C$

For example, what is the integral shortcut for $\int x \, dx$? To determine this, we can use the power rule. In this case, our exponent is $n = 1$. Using the power rule, we have $\int x^1\,dx = \frac{x^{1+1}}{1+1} = \frac{x^{2}}{2} + C$.

Similarly, how do you find the integral of $x^2$? Again, we can use the power rule. In this case, $n = 2$. Using the power rule, we have $\int x^2\,dx = \frac{x^{2+1}}{2+1} = \frac{x^{3}}{3} + C$.

What is the integral of sin(x) dx? For this integral, we can use the trigonometric rules. The sine rule tells us that $\int \sin{(x)} \, dx = -\cos{x} + C$.

2. Use LaTeX to Streamline Formatting

How do you type an integral on a keyboard? The easiest option is to use LaTeX, which is a typesetting system that is optimized to display technical content. You can download LaTeX onto your computer, or you can make an account on a free online LaTeX editor.

To display a definite integral using LaTeX, you can use the code “$\int f(x)\,dx = F(x) + C$.” The dollar signs indicate math mode in LaTeX. To display an indefinite integral, you can use the code “$\int_{a}^{b} f(x)\,dx = F(x)\Big|_a^b$.”

Alternatively, the keyboard shortcuts for Mac and Windows are Option + B and Alt + 244.

3. Integration by Parts Calculus

If the integrand is a product of functions, then integration by parts is the best integral trick to use. Integration by parts reverses the product rule for derivatives. The formula of integration by parts is:

$\int udv = uv - \int vdu$

To use this formula, we can simply separate the integrand into a product of functions by choosing one function to represent $u$ and one function to represent $dv$. It’s best to designate $u$ as the term that’s easiest to differentiate, and $dv$ as the term that’s easiest to integrate.

The next step is to differentiate $u$ to find $du$ and integrate $dv$ to find $v$. Then, all that’s left to do is plug $u$, $v$, and $du$ into the integration by parts formula and solve, simplifying where needed.

How do you know how to assign $u$ and $du$? When deciding which function to appoint as $u$, one handy trick is to use the acronym LIATE:

Logarithmic functions

Inverse trigonometric functions

Algebraic functions

Trigonometric functions

Exponential functions

Identify the different function types that are present in your integrand. The function types that are placed highest in the above list should be prioritized as $u$.

For example, let’s solve $\int \ln{(x)}dx$. Let $u = ln{(x)}$ and $dv=1dx$. Differentiating $u$, we find that $du = \frac{1}{x}dx$. Integrating $dv$, we find that $v = x$. Now, plugging $u$, $v$, and $du$ into the integration by parts formula:

$\int udv = uv - \int vdu$

$\int \ln{(x)} = \ln{(x)} \cdot x - \int x \cdot \frac{1}{x}dx$

$\int \ln{(x)} = x \ln{(x)} - \int 1dx$

$\int x\ln{(x)} = x \ln{(x)} - x + C$

4. U-Substitution

If the integrand is a composite function, then u-substitution is the best integral trick to use. By using u-substitution, we can easily reverse the chain rule for derivatives. To use this trick, we rewrite our integral in terms of $u$ and $du$:

$\int f(g(x))g’(x)\,dx = \int f(u)\,du$

Make sure to replace all forms of $x$ in your substitution. To do this, you might need to algebraically manipulate the problem a little bit; this might involve rewriting an expression in terms of $u$, or dividing or multiplying the integral by a constant. It can be helpful to solve for $dx$ in terms of $du$ to determine if this is the case.

For example, let’s integrate $(7- x)^4$. Let $u = 7 - x$. Then $du = -1dx$, so $dx = -1 du$. After substituting these values into our integrand, we can use the power rule. Remember to substitute the original values back into your final answer.

$\int f(g(x))g’(x)\,dx = \int f(u)\,du$

$\int (7- x)^4 \, dx = \int u^4 (-1)du$

$= - \frac{u^5}{5} + C$

$= - \frac{(7 - x)^5}{5} + C$

5. Trigonometric Identities

If the integrand involves trigonometric functions, one helpful trick is to use trigonometric identities to transform the integrand into a form that is more easily integrated.

Here are some of the most commonly used trigonometric identities:

Pythagorean Identities

$\sin^2x + \cos^2x = 1$

$\sec^2 x - \tan^2 x = 1$

$\csc^2 x - \cot^2 x = 1$

Quotient and Reciprocal Identities

$\tan x = \frac{\sin x}{\cos x}$

$\cot x = \frac{\cos x}{\sin x}$

$\sin x = \frac{1}{\csc x}$

$\cos x = \frac{1}{\sec x}$

$\tan x = \frac{1}{\cot x}$

$\csc x = \frac{1}{\sin x}$

$\sec x = \frac{1}{\cos x}$

$\cot x = \frac{1}{\tan x}$

Double Angle Identities

$\sin{(2x)} = 2 \sin x \cos x$

$\cos{(2x)} = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$

$\tan{(2x)} = \frac{2 \tan x}{1 - \tan^2 x}$

6. Integrating Even and Odd Functions

For some definite integral problems, identifying if a function is even or odd can quickly simplify the problem.

If $f(-x) = f(x)$, the function is even.

If $f(-x) = -f(x)$, the function is odd.

After identifying that one of the above cases is true, examine the integral bounds. Determine if the lower and upper bounds are in the form $-a$ and $a$. If so, then the following tricks can be used:

If $f(x)$ is even, then $\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx$.

If $f(x)$ is odd, then $\int_{-a}^a f(x) \, dx = 0$.

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