This section covers:
WARNING: The techniques in this section only work if the argument of what’s being integrated is just “\(x\)”; in other words, “\(x\)” is by itself and doesn’t have a coefficient or perhaps more complicated. See the USubstitution Integration section for more integrating more complicated expressions!
Antiderivatives (Antiderivatives)
Antiderivatives are just the opposite of derivatives; it’s that simple. For example, since we know that the derivative of \(5x\) is 5, the antiderivative of 5 is \(5x\). But we have to be careful here, since the derivative of \(5x+4\) is also 5 (since the derivative of a constant is 0). So, technically, the antiderivative of 0 is “\(C\)”, where \(C\) is any constant, because the derivative of \(C\) is 0.
When we take the antiderivative of something, we are actually integrating it, or taking the integral of it. The integrand is the name of what we’re taking the integral of.
Here are definitions of an antiderivative, integral, constant of integration, and differential equation. We’ll talk more about Differential Equations in the Differential Equations and Slope Fields section.
Definitions of Antiderivative (Integral):
A function \(F\) is an antiderivative of another function \(f\) when \({F}’\left( x \right)=f\left( x \right)\).
Note that the term indefinite integral is another word for an antiderivative, and is denoted by the integral sign \(\int{.}\)
When we have the differential equation (an equation that involve \(x\), \(y\) and the derivative of \(y\)) in the form \(\displaystyle \frac{{dy}}{{dx}}=f\left( x \right)\), we can write it as \(dy=f\left( x \right)dx\). When we integrate, we have \(\displaystyle y=\int{{f\left( x \right)dx}}=F\left( x \right)+\,\,C\), where \(C\) is the constant of integration.
But the most important thing to remember is that integration is the opposite of differentiation!
Basic Integration Rules
I know this seems confusing at this point, but let’s go through some basic integration rules and examples, and then do some problems!
Differentiation Formula  Integration Formula  Examples 
\(\displaystyle \frac{d}{{dx}}\left( \text{C} \right)=0\) (\(C\) is a constant)  \(\int{0}\,\,dx=\text{C}\)  \(\displaystyle \frac{d}{{dx}}\left( \text{6} \right)=0\)
\(\int{0}\,\,dx\text{ can equal 6}\) 
\(\displaystyle \frac{d}{{dx}}\left( {kx} \right)=k\) (\(k\) is a constant)  \(\int{k}\,dx=kx+C\)  \(\displaystyle \frac{d}{{dx}}\left( {3x} \right)=3\)
\(\int{3}\,dx=3x+C\) 
\(\displaystyle \frac{d}{{dx}}\left( {k\cdot f\left( x \right)} \right)=k\cdot {f}’\left( x \right)\)  \(\int{{k\cdot f\left( x \right)dx}}=k\int{{f\left( x \right)dx}}\)  \(\displaystyle \frac{d}{{dx}}\left( {4x} \right)=4\cdot \text{derivative of }x=4\cdot 1=4\)
\(\int{{4\cdot 1\,dx}}=4\int{{\left( 1 \right)dx}}=4x\) 
\(\displaystyle \frac{d}{{dx}}\left( {f\left( x \right)\,\pm \,g\left( x \right)} \right)\)
\(\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,={f}’\left( x \right)\,\,\pm \,\,{g}’\left( x \right)\) 
\(\begin{array}{l}\int{{\left( {f\left( x \right)\,\,\pm \,\,g\left( x \right)} \right)}}\,\,dx=\\\,\,\,\,\,\,\,\,\int{{f\left( x \right)dx\pm \,\,\int{{g\left( x \right)}}}}\,dx\end{array}\)  \(\displaystyle \frac{d}{{dx}}\left( {4x+3x} \right)={f}’\left( {4x} \right)\,+\,{g}’\left( {3x} \right)=4+3=7\)
\(\int{{\left( {4+3} \right)}}dx=\int{{4\,dx+\int{3}}}dx=4x+3x\) 
\(\displaystyle \frac{d}{{dx}}\left( {{{x}^{n}}} \right)=n{{x}^{{n1}}}\)  \(\displaystyle \int{{{{x}^{n}}dx}}=\frac{{{{x}^{{n+1}}}}}{{n+1}};\,\,n\ne 1\)  \(\displaystyle \frac{d}{{dx}}\left( {{{x}^{3}}} \right)=3{{x}^{{31}}}=3{{x}^{2}}\)
\(\require {cancel} \displaystyle \int{{3{{x}^{2}}dx}}=3\int{{{{x}^{2}}dx}}=3\cdot \frac{{{{x}^{{2+1}}}}}{{2+1}};=\cancel{3}\cdot \frac{{{{x}^{3}}}}{{\cancel{3}}}={{x}^{3}}\)

Trigonometric Integration Rules
Here are the trigonometry integration rules:
Trig Differentiation Formula  Trig Integration Formula 
\(\displaystyle \frac{d}{{dx}}\left( {\sin x} \right)=\cos x\)  \(\int{{\cos x}}dx=\sin x+C\) 
\(\displaystyle \frac{d}{{dx}}\left( {\cos x} \right)=\sin x\)  \(\int{{\sin x}}dx=\cos x+C\) 
\(\displaystyle \frac{d}{{dx}}\left( {\tan x} \right)={{\sec }^{2}}x\)  \(\int{{{{{\sec }}^{2}}\,x}}dx=\tan x+C\) 
\(\displaystyle \frac{d}{{dx}}\left( {\sec x} \right)=\sec x\tan x\)  \(\int{{\sec x\tan x}}dx=\sec x+C\) 
\(\displaystyle \frac{d}{{dx}}\left( {\cot x} \right)={{\csc }^{2}}x\)  \(\int{{{{{\csc }}^{2}}x}}dx=\cot x+C\) 
\(\displaystyle \frac{d}{{dx}}\left( {\csc x} \right)=\csc x\cot x\)  \(\int{{\csc x\cot x}}dx=\csc x+C\) 
Here are some hints to help you remember the trig differentiation and integration rules:
When the trig functions start with “c”, the differentiation or integration is negative (cos and csc). For the functions other than sin and cos, there’s always either one tan and two secants, or one cot and two cosecants on either side of the formula. Look at the formulas and see how this makes sense!
Note: More trig integration rules (involving “ln”) will be introduced later here in the Exponential and Logarithmic Integration section.
Indefinite Integration Problems
Let’s do some problems; notice how we may need to rewrite the integral, and then simplify at the end. Note to check your work, you can differentiate back from the answer to see if you get the original!
Also, remember that you can check these by going to www.wolframalpha.com (or use app on smartphone) and type in “integral of x^.25” for example.
Original Integral  Rewritten Integral  Integrated Answer  Simplified Answer 
\(\int{{\sqrt[4]{x}}}dx\)  \(\int{{{{x}^{{\frac{1}{4}}}}}}dx\)  \(\displaystyle \frac{{{{x}^{{\frac{1}{4}+1}}}}}{{\frac{1}{4}+1}}+C=\frac{{{{x}^{{\frac{5}{4}}}}}}{{\frac{5}{4}}}+C\)  \(\displaystyle \frac{4}{5}{{x}^{{\frac{5}{4}}}}+C\) 
\(\displaystyle \int{{\frac{1}{{2{{x}^{5}}}}}}dx\)  \(\displaystyle \frac{1}{2}\int{{{{x}^{{5}}}}}dx\)  \(\displaystyle \frac{1}{2}\cdot \frac{{{{x}^{{5+1}}}}}{{5+1}}+C=\frac{1}{2}\cdot \frac{{{{x}^{{4}}}}}{{4}}+C\)  \(\displaystyle \frac{1}{{8{{x}^{4}}}}+C\) 
\(\displaystyle \int{{\frac{5}{{{{{\left( {3x} \right)}}^{3}}}}}}dx\)  \(\displaystyle \int{{\frac{5}{{27{{x}^{3}}}}}}dx=\frac{5}{{27}}\int{{{{x}^{{3}}}}}dx\)  \(\displaystyle \frac{5}{{27}}\cdot \frac{{{{x}^{{3+1}}}}}{{3+1}}+C=\frac{5}{{27}}\cdot \frac{{{{x}^{{2}}}}}{{2}}+C\)  \(\displaystyle \frac{5}{{54{{x}^{2}}}}+C\) 
\(\displaystyle \int{{\frac{{2\sqrt[3]{x}}}{{{{x}^{2}}\sqrt{x}}}}}dx\)  \(\begin{array}{c}2\int{{{{x}^{{\frac{1}{3}}}}\cdot {{x}^{{2}}}\cdot {{x}^{{\frac{1}{2}}}}dx}}\\=2\int{{{{x}^{{\frac{{13}}{6}}}}dx}}\end{array}\)
(add exponents when multiplying!) 
\(\displaystyle 2\cdot \frac{{{{x}^{{\frac{{13}}{6}+1}}}}}{{\frac{{13}}{6}+1}}+C=2\cdot \frac{{{{x}^{{\frac{7}{6}}}}}}{{\frac{7}{6}}}+C\)  \(\displaystyle \frac{{12}}{{7{{x}^{{\frac{7}{6}}}}}}+C\)
\(\displaystyle =\frac{{12}}{{7{{x}^{7}}\sqrt[6]{x}}}+C\)

\(\int{{4\sin x}}dx\)  \(4\int{{\sin x}}dx\)  \(4\cdot \cos x+C\)  \(4\cos x+C\) 
Integration Problem  Solution 
\(\int{{\left( {{{x}^{2}}+5x+1} \right)}}dx\) 
\(\displaystyle \int{{\left( {{{x}^{2}}+5x+1} \right)}}dx=\int{{{{x}^{2}}dx+5\int{x}dx+\int{1}\,dx}}=\frac{{3{{x}^{2}}}}{3}+\frac{{5{{x}^{2}}}}{2}+x+C\) Check: \(\displaystyle \frac{d}{{dx}}\left( {\frac{{{{x}^{3}}}}{3}+\frac{{5{{x}^{2}}}}{2}+x+C} \right)=\frac{1}{3}\cdot 3{{x}^{2}}+\frac{1}{2}\cdot 10x+1={{x}^{2}}+5x+1\,\,\surd \) 
\(\displaystyle \int{{\left( {\frac{{\sqrt{x}+5}}{{{{x}^{2}}}}} \right)}}dx\) 
\(\displaystyle \int{{\left( {\frac{{\sqrt{x}+5}}{{{{x}^{2}}}}} \right)}}dx=\int{{\frac{{{{x}^{{\frac{1}{2}}}}}}{{{{x}^{2}}}}}}dx+\int{{\frac{5}{{{{x}^{2}}}}}}dx=\int{{{{x}^{{\frac{3}{2}}}}}}dx+5\int{{{{x}^{{2}}}}}dx\) \(\displaystyle =\frac{{{{x}^{{\frac{1}{2}}}}}}{{\frac{1}{2}}}+\frac{{5{{x}^{{1}}}}}{{1}}+C=\frac{2}{{\sqrt{x}}}\frac{5}{x}+C=\frac{{2\sqrt{x}+5}}{x}+C\) Check: \(\displaystyle \frac{d}{{dx}}\left( {\frac{{2\sqrt{x}+5}}{x}} \right)=\frac{d}{{dx}}\left( {2{{x}^{{\frac{1}{2}}}}5{{x}^{{1}}}+C} \right)\) \(\displaystyle =1\cdot {{x}^{{\frac{3}{2}}}}\left( {5{{x}^{{2}}}} \right)={{x}^{{\frac{3}{2}}}}+5{{x}^{{2}}}\) \(\displaystyle =\frac{{{{x}^{{\frac{1}{2}}}}}}{{{{x}^{2}}}}+\frac{5}{{{{x}^{2}}}}=\frac{{\sqrt{x}+5}}{{{{x}^{2}}}}\,\,\surd \) 
\(\int{{\left( {4{{{\sec }}^{2}}\theta +\cos \theta } \right)}}d\theta \) 
\(\int{{\left( {4{{{\sec }}^{2}}\theta +\cos \theta } \right)}}d\theta =4\int{{{{{\sec }}^{2}}\theta }}d\theta +\int{{\cos \theta }}d\theta =4\tan \theta +\sin \theta +C\) Check: \(\displaystyle \frac{d}{{d\theta }}\left( {4\tan \theta +\sin \theta +C} \right)=4{{\sec }^{2}}\theta +\cos \theta \,\,\surd \) 
Initial Conditions and Particular Solutions
Since when we take an integral of a function, we have to add the “\(+C\)” part to it (the constant of integration), if we were to graph the solutions, we’d have many different curves that are the same but vertical translations of each other. But sometimes we know more about the curve, so we can determine a particular solution (get what the \(C\) is), and we can do this say with one \((x,y)\) value on the curve, which is called the initial condition.
So basically what you do is take the integral with the “\(+C\)” part, and then put in the initial condition values given for \(x\) and \(y\) (which is \(f(x)\)), and solve for \(C\). This will give you the particular solution.
Note that we have to go through the exercise more than once when we are given the second derivative.
Let’s do some examples:
And here’s a word problem. Note how in this problem, we have to use a system of equations to solve for the particular solution. Also remember when something is proportional to something else, it’s a direct variation, and one side is the product of a constant \(k\) and the other side.
Revisiting Position, Velocity, and Acceleration
We talked a little about Rates of Change and Velocity here in the Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change section (and we will here in the Definite Integration section). Now we’ll be able to work backwards (for example, from a rate to a distance) since we know how to integrate functions!
Here are some things we know about position, velocity, and acceleration:
 If a particle is moving along a horizontal line, its position (relative to say the origin) is a function, but the derivative of this function would be its (instantaneous) velocity (how fast it’s moving) at a certain point, and the derivative of its velocity would be its acceleration (how fast its velocity is changing).
 We can take the integral of velocity to get position, and the integral of acceleration to get velocity.
 The position of an object is actually a vector, since it has both a magnitude (a scalar, such as distance) and a direction. A change in position is a displacement, which is how far out of place the object is, compared to where it started. The distance it has traveled is the total amount of ground an object has covered during its motion; this is the absolute value of the displacement, and is a scalar.
 The velocity function is the derivative of the position function, and be negative, zero, or positive. If the derivative (velocity) is positive, the object is moving to the right (or up, if that’s how the coordinate system is defined); if negative, it’s moving to the left (or down); if the velocity is 0, the object is at rest. This is also called the direction of the object.
 The velocity of an object is actually a vector, whereas the speed is the absolute value of the velocity, and is a scalar. The speed of an object cannot be negative, whereas velocity can.
 Acceleration (the derivative of velocity, which is also a vector) can cause speed to increase, decrease, or stay the same. Negative acceleration means slowing down (velocity decreasing) and positive acceleration means speeding up (velocity increasing).
Let’s do some problems:
Here’s one more problem:
Understand these problems, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
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On to USubstitution Integration – you’re ready!