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Best Integral Shortcuts: Solve & Master Integrals

09.26.2022 • 8 min read

Rachel McLean

Subject Matter Expert

Learn how to integrate quickly with these integration shortcuts.


In This Article

  1. What Is an Integral?

  2. Top 6 Integral Shortcuts

  3. Practice Makes Perfect

What Is an Integral?

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 yy, and each rectangle’s width is a tiny change in xx. This gives us the definition of a definite integral, where AA represents the area under the curve on [a,b][a, b].

abf(x)dx=A\int_{a}^{b} f(x)\,dx = A
Graph showing Second Fundamental Theorem of Calculus

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

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

abf(x)dx=F(x)ab=F(b)F(a)\int_{a}^{b} f(x)\,dx = F(x)\Big|_a^b = F(b) - F(a)

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

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

If F(x)=f(x)F’(x) = f(x), then F(x)F(x) is an antiderivative of f(x)f(x). This means that taking the derivative of F(x)F(x) gives back f(x)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 f(x)dx=F(x)+C\int f(x)\,dx = F(x) + C, then F(x)=f(x)F’(x) = f(x). As mentioned earlier, the antiderivative function F(x)F(x) can be used to calculate all possible areas under the curve by taking the difference of F(x)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:

f(x)dx=f(x)+C\int f’(x)\,dx = f(x) + C

CC represents the constant of integration. Since the derivative of any constant is zero, CC 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

These equations are given below.

Sum Rule

[f(x)+g(x)]dx=f(x)dx+g(x)dx\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx

Difference Rule

[f(x)g(x)]dx=f(x)dxg(x)dx\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx

Constant Multiplier Rule

kf(x)dx=kf(x)dx\int kf(x)\,dx = k\int f(x)\,dx for some constant kk

Power Rule

xndx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C for some real number nn

Constant Rule

adx=ax+C\int a\,dx = ax + C for some constant aa

Dr. Tim Chartier, an Outlier instructor of Calculus, explains several of these rules:

4 Integral Formulas for Specific Function Types

Once you’re familiar with the basic rules above, you can memorize these integral formulas for specific function types:

Reciprocal Rules

  • 1xdx=x1dx=lnx+C\int \frac{1}{x}\,dx = \int x^{-1}\,dx = \ln{|x|} + C

  • 1ax+bdx=1aln(ax+b)+C\int\frac{1}{ax+b}\,dx = \frac{1}{a} \ln{(ax+b)} + C

Exponential Functions & Logarithmic Functions Rules

  • exdx=ex+C\int e^x\,dx = e^x + C

  • axdx=axln(a)+C\int a^x\,dx = \frac{a^x}{\ln{(a)}} + C, for any positive real number aa

  • ln(x)dx=xln(x)x+C\int \ln{(x)}\,dx = x\ln{(x)}-x + C

Trigonometric Functions & Inverse Trigonometric Functions Rules

For these rules, assume that x is in radians.

  • sinxdx=cosx+C\int \sin{x}\,dx = -\cos{x} + C

  • cosxdx=sinx+C\int \cos{x}\,dx = \sin{x} + C

  • sec2xdx=tanx+C\int \sec ^2 x\,dx = \tan{x} + C

  • csc2xdx=cotx+C\int \csc ^2 x\,dx = -\cot{x} + C

  • secxtanxdx=secx+C\int \sec{x}\tan{x}\,dx = \sec{x} + C

  • cscxcotxdx=cscx+C\int \csc{x}\cot{x}\,dx = -\csc{x} + C

  • dx1x2=sin1x+C\int \frac{dx}{\sqrt{1-x^2}}=\sin ^{-1}x + C

  • dx1x2=cos1x+C\int \frac{-dx}{\sqrt{1-x^2}}=\cos ^{-1}x + C

  • dx1+x2=tan1x+C\int \frac{dx}{1+x^2}=\tan ^{-1}x + C

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

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

Absolute Value Rule

xdx=xx2+C\int |x|\,dx = \frac{x |x|}{2} + C

For example, what is the integral shortcut for xdx\int x \, dx? To determine this, we can use the power rule. In this case, our exponent is n=1n = 1. Using the power rule, we have x1dx=x1+11+1=x22+C\int x^1\,dx = \frac{x^{1+1}}{1+1} = \frac{x^{2}}{2} + C.

Similarly, how do you find the integral of x2x^2? Again, we can use the power rule. In this case, n=2n = 2. Using the power rule, we have x2dx=x2+12+1=x33+C\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 sin(x)dx=cosx+C\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 “f(x)dx=F(x)+C\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 “abf(x)dx=F(x)ab\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:

udv=uvvdu\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 uu and one function to represent dvdv. It’s best to designate uu as the term that’s easiest to differentiate, and dvdv as the term that’s easiest to integrate.

The next step is to differentiate uu to find dudu and integrate dvdv to find vv. Then, all that’s left to do is plug uu, vv, and dudu into the integration by parts formula and solve, simplifying where needed.

How do you know how to assign uu and dudu? When deciding which function to appoint as uu, 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 uu .

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

udv=uvvdu\int udv = uv - \int vdu
ln(x)=ln(x)xx1xdx\int \ln{(x)} = \ln{(x)} \cdot x - \int x \cdot \frac{1}{x}dx
ln(x)=xln(x)1dx\int \ln{(x)} = x \ln{(x)} - \int 1dx
xln(x)=xln(x)x+C\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 uu and dudu :

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

Make sure to replace all forms of xx 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 uu , or dividing or multiplying the integral by a constant. It can be helpful to solve for dxdx in terms of dudu to determine if this is the case.

For example, let’s integrate (7x)4(7- x)^4 . Let u=7xu = 7 - x . Then du=1dxdu = -1dx , so dx=1dudx = -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.

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)\,dx = \int f(u)\,du
(7x)4dx=u4(1)du\int (7- x)^4 \, dx = \int u^4 (-1)du
=u55+C= - \frac{u^5}{5} + C
=(7x)55+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

  • sin2x+cos2x=1\sin^2x + \cos^2x = 1

  • sec2xtan2x=1\sec^2 x - \tan^2 x = 1

  • csc2xcot2x=1\csc^2 x - \cot^2 x = 1

Quotient and Reciprocal Identities

  • tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}

  • cotx=cosxsinx\cot x = \frac{\cos x}{\sin x}

  • sinx=1cscx\sin x = \frac{1}{\csc x}

  • cosx=1secx\cos x = \frac{1}{\sec x}

  • tanx=1cotx\tan x = \frac{1}{\cot x}

  • cscx=1sinx\csc x = \frac{1}{\sin x}

  • secx=1cosx\sec x = \frac{1}{\cos x}

  • cotx=1tanx\cot x = \frac{1}{\tan x}

Double Angle Identities

  • sin(2x)=2sinxcosx\sin{(2x)} = 2 \sin x \cos x

  • cos(2x)=cos2xsin2x=2cos2x1=12sin2x\cos{(2x)} = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x

  • tan(2x)=2tanx1tan2x\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)f(-x) = f(x) , the function is even.

If f(x)=f(x)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-a and aa . If so, then the following tricks can be used:

If f(x)f(x) is even, then aaf(x)dx=20af(x)dx\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx.

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

Practice Makes Perfect

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