Explanations 
Flashcards 
StudySmarter AI 
Notes 
Study Plans 
Study Sets 
Exams 
Magazine 
Find a job 
Mobile App 
Jump to a key chapter
Teacher: What is 3+3?
Bobby: 3!
Teacher: Yes, Bobby, that is correct.
Jokes aside, you have probably come across factorials in a variety of situations. Normally, we define factorials by writing:
$$n! = n(n-1)...1$$
This definition makes sense if n is a positive integer, but not if \(n\) is any other type of real number. Mathematicians, being mathematicians, decided that this state of things was, frankly, unacceptable. So, they came up with definitions of factorials that allow you to do things like find \(\pi!\). One of these is the gamma function
$$n! = \Gamma(n+1) = \int_0^{\infty} x^{n-1}e^{-x} \; dx.$$
This is great, but presents us with another dilemma: how do we evaluate this (frankly somewhat scary) integral? Fortunately, there are a variety of techniques of integration we can throw at the problem. The most common techniques of integration are:
The Power Rule for Integration
Integration by Substitution
Integration by Parts
Integration by Partial Fractions
Integrating Functions Using Long Division
These are essential to know, but won't always help you with integrals like the gamma function. There are too many integration techniques for any one article to cover fully, but here you will see a few of the better-known techniques for dealing with problematic integrals. In particular, this article will cover the power rule for integration, integrals of inverse functions, Weierstrass substitution, and Feynman's technique of integration. For more details on the other techniques of integration listed above, see the corresponding articles.
Techniques of Integration in Calculus
The Power Rule
Just like there is a power rule for differentiation, there is a power rule for integration.
The Power Rule for Integration states that
$$\displaystyle\int ax^n \; dx = \frac{a}{n+1}x^{n+1} + C.$$
Here, \(n\) can be any real number (positive, negative, zero, integer, rational, or irrational).
We can prove the power rule for integration directly from the power rule for differentiation. Recall that the power rule for differentiation states that
$$\frac{d}{dx}x^n = nx^{n-1}.$$
Given a function
$$\frac{a}{n+1}x^{n+1},$$
using the power rule for differentiation, we know that
$$\begin{align}\frac{d}{dx}\frac{a}{n+1}x^{n+1} &= \frac{a}{n+1}\frac{d}{dx} x^{n+1}\\ &= \frac{a}{n+1}\left((n+1)x^n\right)\\ &= ax^n.\end{align}$$
Integrating both sides of the equation and using the Fundamental Theorem of Calculus, we get that
\[\int \frac{d}{dx}\frac{a}{n+1}x^{n+1}\;dx = \int ax^n \; dx. \]
or in other words
\[ \frac{a}{n+1}x^{n+1} + C = \int ax^n \; dx.\]
You can think of the power rule for integration as 'undoing' the power rule for differentiation. Many other common integration techniques are based on this strategy of 'undoing' differentiation rules.
Evaluate the integral
\[\int 3x^3 \; dx.\]
Solution:
Here, you can use the Power Rule for Integration with \(a=3\) and \(n=3\). So,
\[\int 3x^3 \; dx = \frac{3}{4}x^{4} + C.\]
You can check by differentiating that this answer is correct.
Other Common Techniques
Some common techniques of integration include:
Integration by Substitution
Integration by Parts
Integration by Partial Fractions
Integrating Functions Using Long Division
For more information on each of these, see the corresponding articles.
Examples of Common Integration Techniques
The Power Rule
Let's do a few examples using the Power Rule for Integration.
First, we can use the Power Rule to evaluate integrals with added terms and radicals.
Evaluate the integral
\[\int \sqrt{x} + \frac{17}{\sqrt{\pi}}x^{16} \; dx .\]
Solution:
The first thing to notice is that you have two terms in our integral added to each other. By the sum rule for integration, you can write
\[\int \sqrt{x} + \frac{17}{\sqrt{\pi}}x^{16} \; dx = \int \sqrt{x} \; dx + \int \frac{17}{\sqrt{\pi}}x^{16} \; dx.\]
The second term can be integrated by directly applying the power rule with \(a = \frac{17}{\sqrt{\pi}}\) and \(n = 16\). To integrate the first term, use the identity
\[\sqrt{x} = x^{1/2}.\]
Using this identity and applying the power rule, you get that
\[\begin{align}\int \sqrt{x} + \frac{17}{\sqrt{\pi}}x^{16} \; dx &= \int \sqrt{x} \; dx + \int \frac{17}{\sqrt{\pi}}x^{16} \; dx\\&= \int x^{1/2} \; dx + \int \frac{17}{\sqrt{\pi}}x^{16} \; dx\\&= \frac{1}{3/2}x^{3/2} + \frac{17/\sqrt{\pi}}{17}x^{17}\\&= \frac{2}{3}x^{3/2} + \frac{1}{\sqrt{\pi}}x^{17}.\end{align}\]
Let's do another example, this time using negative exponents.
Evaluate
\[\int \frac{1}{x^3} - 4x^{3/2} \; dx.\]
Solution:
Since the integrand is the difference of two terms, you can split up the integral using the difference rule for integrals:
\[\int \frac{1}{x^3} - 4x^{3/2} \; dx = \int \frac{1}{x^3} \; dx - \int 4x^{3/2} \; dx\]
To integrate the first term, use the identity
\[\frac{1}{x^3} = x^{-3}.\]
This gives you:
\[\begin{align}\int \frac{1}{x^3} - 4x^{3/2} \; dx &= \int \frac{1}{x^3} \; dx - \int 4x^{3/2} \; dx\\&= \int x^{-3} dx - \int 4x^{3/2} \; dx\\&= \frac{1}{-2}x^{-2} - \frac{4}{5/2} x^{5/2}\\&= -\frac{1}{2}x^{-2} - \frac{8}{5} x^{5/2}.\end{align}\]
Additional Techniques of Integration
Finding new techniques of integration is almost a cottage industry among mathematicians. This is because techniques of integration can be interesting, unexpected, and just flat-out fun to mess with. It is beyond the scope of this article to detail every technique of integration known, but we can look at a few examples. In particular, we will look at integrals of inverse functions, techniques for trigonometric integrals, and Feynman's technique of integration.
Integrals of Inverse Functions
Indefinite Integrals
There is a nice result that gives the integral of any inverse function. (For a refresher, see the article Inverse Functions.)
\[\int f^{-1}(y) dy = yf^{-1}(y) - F(f^{-1}(y)) + C.\]
Note that \(f(x)\) that has inverse \(f^{-1}(x)\) and antiderivative \(F(x)\) means that \(\frac{d}{dx}F(x) = f(x)\) .
Here, \(y\) is used instead of \(x\) to emphasize that you are working with inverse functions. This equation can be directly verified by differentiating, using the identity
\[\frac{d}{dy}f^{-1}(y) = \frac{1}{f'(f^{-1}(y))}. \]
With the identity, you get
\[\begin{align}\frac{d}{dy}\left(yf^{-1}(y) - F(f^{-1}(y)) + C\right) &= \frac{d}{dy}\left(yf^{-1}(y)\right) - \frac{d}{dy}\left(F(f^{-1}(y)) + C\right)\\&= f^{-1}(y) + y\left(\frac{1}{f'(f^{-1}(y))}\right) - F'(f^{-1}(y))\frac{1}{f'(f^{-1}(y))}\\&= f^{-1}(y) + y\left(\frac{1}{f'(f^{-1}(y))}\right) - f(f^{-1}(y))\frac{1}{f'(f^{-1}(y))}\\&= f^{-1}(y) + y\left(\frac{1}{f'(f^{-1}(y))}\right) - y\frac{1}{f'(f^{-1}(y))}\\&= f^{-1}(y),\end{align}\]
which is exactly what you wanted to see.
Definite Integrals
There is also a version of this theorem for definite integrals.
Given a function f with inverse \(f^{-1}\),
\[\int_{f(a)}^{f(b)} f^{-1}(y) dy + \int_a^b f(x) \; dx = bf(b)-af(a).\]
There is a nice visual proof of this fact using properties of inverse functions. First, for simplicity's sake, set \(a = f(a) = 0\) and \(b > 0, \, f(b) > 0\). You can draw the integral \(\int_0^f(b) f(x) \; dx\) like this, where the shaded area is the value of the integral:
Fig. 1. Integral of a function as an area.
You might remember that reflecting the graph of \(f\) about the line \(y=x\) gives the graph of \(f^{-1}\):
Fig. 2. Symmetry of a function and its inverse across the line\(y=x\).
You need to find the sum of the integral of the function and its inverse. This is where you can use a nice trick: treat \( f^{-1} \) as a function of \(y\), not as a function of \(x\). Algebraically, this would mean replacing every occurrence of \(x\) with \(y\) and every \(y\) with an \(x\) in the equation for \(f^{-1}\). This is just a change of variables; it doesn't change the integral you are working with at all. Here's the kicker: replacing \(y\) with \(x\) and \(x\) with \(y\) is geometrically equivalent to flipping the graph of \(f^{-1}\) about the line \(y=x\), as you can see in the picture below.
Fig. 3. Visual proof of inverse function integral as an area.
So, as the picture above shows, the sum of our integrals is the same as the area of the rectangle, which is \(bf(b)\).
Now, if you let \(a\) and \(c\) be non-zero, all you are doing geometrically is cutting out a rectangle with area \(af(a)\) from the area you are looking for as you can see in the graph below.
Fig. 4. Visual proof of the integral of an inverse function with a non-zero limit of integration.
Thus, in general,
\[\int_{f(a)}^{f(b)} f^{-1}(y)dy + \int_a^b f(x) \; dx = bf(b)-af(a).\]
Examples
Let's do an example of finding the antiderivative of an inverse function.
Evaluate
\[\int \cos^{-1}(y)dy\]
where \(\cos(x)\)) is considered as a function on the interval \([0,\pi].\)
Solution:
The function \(\cos^{-1}(y)\) is the inverse of the function \(\cos(x)\), so you can use the formula
\[\int f^{-1}(y) dy = yf^{-1}(y) - F(f^{-1}(y)) + C.\]
First, note that
\[F(x) = \int \cos(x) \; dx = -\sin(x) + C.\]
So, plugging the functions into the equation above, you get that
\[\int cos^{-1}(y) \; dx = y\cos^{-1}(y) - \sin(\cos^{-1}(y)) + C.\]
You can simplify this expression somewhat by using properties of trigonometric functions. Let \(\theta = \cos^{-1}(x)\). Then
\[\cos(\theta) = x = \frac{x}{1}.\]
As you recall, the cosine of an angle theta can be interpreted in terms of right triangles as the ratio of the adjacent side of the triangle to its hypotenuse.
Fig. 5. Triangle with angle \(\theta = \cos^{-1}(y)\).
Using this same triangle, you get that
\[\sin(\cos^{-1}(y)) = \sin(\theta) = \sqrt{1-y^2}.\]
Thus, the expression simplifies to
\[\int cos^{-1}(y) \; dx = y\cos^{-1}(y) - \sqrt{1-y^2} + C.\]
This 'triangle trick' is handy for many integrals. See the article Trigonometric Substitution for more examples.
Let's do an example of finding the definite integral of an inverse function.
Evaluate
\[\int_1^e \ln(y) dy = \int_{e^0}^{e^1} \ln(y) dy.\]
Solution:
First, note that \(\ln(y)\) is the inverse of \(e^x\). The integral of \(\ln(y)\) is not necessarily obvious, but we know how to integrate \(e^x\). So, this is a great situation to use properties of inverse functions. Plugging this information into the equation for the definite integral of an inverse function, you get that
\[\begin{align} \int_{e^0}^{e^1} \ln(y) dy + \int_{0}^1 e^x \; dx &= 1(e^1) - 0(e^0) \\ &= e. \end{align}\]
Next, you can integrate to find
\[\begin{align}\int_0^1 e^x \; dx &= e^x \bigg|_{x=0}^{x=1}\\&= e^1 - e^0\\&= e-1.\end{align}\]
Finally, you can solve for
\[\int_{e^0}^{e^1} \ln(y) \; dy. \]
Substituting in what you know,
\[ \int_{e^0}^{e^1} \ln(y) \; dy + e-1 = e, \]
which means that\[ \int_{e^0}^{e^1} \ln(y) \; dy = 1. \]
Trigonometric Techniques of Integration
Trigonometric functions turn up in many integrals and can be quite useful, even in unexpected places. For details on trig substitution or how to integrate trig functions in general, see the articles Trigonometric Substitution, Trigonometric Integrals, and Integrals Resulting in Inverse Trigonometric Functions. Here, you can take a look at Weierstrass Substitution, an interesting technique used to evaluate rational functions of sine and cosine. This technique relies on \(u\) substitution, so it may be helpful to read the article Integration by Substitution before reading this section.
Weierstrass Substitution
Weierstrass substitution is an elegant method of solving integrals that are rational functions of sine and cosine. The Weierstrass substitution is the substitution \(u = \tan(x/2)\). Using double angle identities, this substitution gives us the formulas:
\[\begin{align} \sin(x) &= \frac{2u}{1+u^2}, \\ \cos(x)&=\frac{1-u^2}{1+u^2}, \\ dx &= \frac{2}{1+u^2} \; du. \end{align}\]
This technique is particularly useful when sine or cosine functions are in the denominator of an integral.
Use Weierstrass substitution to find
\[\int \frac{dx}{\cos(x)}.\]
Solution:
First, make the substitution
\[ \begin{align} u &= \tan\left(\frac{x}{2}\right),\\ du &= \frac{1+u^2}{2}dx. \end{align}\]
You can use the equations
\[\sin(x) = \frac{2u}{1+u^2}\]
and
\[dx = \frac{2}{1+u^2} \; du\]
to get that
\[\begin{align}\int \frac{1}{\sin(x)}dx &= \int \frac{1+u^2}{2u}\left(\frac{2}{1+u^2}\right) \; du\\&= \int \frac{1}{u} \; du\\&= \ln|u| + C\\&= \ln\left|\tan\left(\frac{x}{2}\right)\right| + C.\end{align}\]
Feynman's Technique of Integration
Feynman's integration technique is an interesting integration technique that is sometimes also referred to as 'differentiating under the integral sign'. Feynman's technique allows you to use differentiation on complicated integrals to obtain an expression that is (hopefully!) easier to integrate.
Steps in Feynman's Technique of Integration
Feynman's technique is difficult to express succinctly because of how much it varies between different integrals. However, the following steps give at least an outline of what we mean by 'Feynman's technique' for evaluating
\[\int_a^b f(x) dx.\]
Define a function\[I(t) = \int_a^b f(x, t)\]by adding a term \(t\) to the integral. You want to make sure that \(I(0) = 0\) and\[I(c) = \int_a^b f(x) dx\]for some \(c\). The best \(f(x, t)\) to use varies widely depending on which integral you are working with.
For example, if your integral was\[ \int_a^b f(x) dx = \int_a^b x^2 dx,\] you might define\[I(t) = \int_a^b x^t dx= \int_a^b f(x, t) dx.\]
Find \(I'(t)\) by differentiating with respect to \(t\) under the integral sign.
In symbols,\[ I'(t) = \frac{d}{dt}\int_a^b f(x, t) dx = \int_a^b \frac{\partial}{\partial t} f(x, t) dx.\]
Using the Fundamental Theorem of Calculus, integrate \(I'(t)\) to find \(I(c)\).
In symbols,\[I(c) = \int_0^c I'(t) dt = \int_a^b f(x) dx\]
Here, the expression \(\frac{\partial}{\partial t}\) just means 'differentiate the expression with respect to \(t\), not with respect to \(x\).' This is called a partial derivative; you will see more of these if you take multivariable calculus.
Examples of Feynman's Technique
Let's do a couple of examples to illustrate this technique.
Evaluate the integral
\[\int_0^1 \frac{x^3 - 1}{\ln(x)} dx.\]
Solution:
To evaluate this integral, you first make the perhaps counterintuitive step of introducing a function of a variable \(t\):
\[I(t) = \int_0^1 \frac{x^t - 1}{\ln(x)}dx.\]
By definition,
\[I(3) = \int_0^1 \frac{x^3 - 1}{\ln(x)},\]
which is the original integral. So, your problem now becomes finding \(I(3)\). To achieve this, start by finding the derivative of our function \(I\):
\[\begin{align}I'(t) &= \frac{d}{dt}\int_0^1 \frac{x^t - 1}{\ln(x)} dx\\&= \int_0^1 \frac{\partial}{\partial t}\frac{x^t - 1}{\ln(x)} dx\\&= \int_0^1 \frac{\partial}{\partial t}\left(\frac{x^t}{\ln(x)} - \frac{1}{\ln(x)}\right) dx\\&= \int_0^1 \frac{\ln(x)x^t}{\ln(x)} dx\\&= \int_0^1 x^t dx\\&= \frac{1}{t+1}x^t \bigg|_{x=0}^{x=1}\\&= \frac{1}{t+1}.\end{align}\]
Here, since you are differentiating with respect to \(t\), you can treat \(x\) as a constant.
Next, since
\[\begin{align} I(0) = \int_0^1 \frac{x^0 - 1}{\ln(x)}dx \\ &= \int_0^1 \frac{1-1}{\ln(x)}dx \\ &= \int_0^1 0 dx \\ &= 0,\end{align}\]
you can use The Fundamental Theorem of Calculus to write:
\[\begin{align}I(3) &= I(3) - I(0)\\&= \int_0^3 I'(t) dt\\&= \int_{0}^3 \frac{1}{t+1} dt\\&= \ln(t+1)\bigg|_{t=0}^{t=3}\\&= \ln(4) - \ln(1)\\&= \ln(4).\end{align}\]
Thus,
\[\int_0^1 \frac{x^3 - 1}{\ln(x)} dx = \ln(4).\]
Let's do one more example.
Evaluate
\[\int_0^{\infty} \frac{\tan^{-1}(x)}{x(1+x^2)} \; dx. \]
See the article Improper Integrals for information on how to solve integrals of this form.
Solution:
First, introduce a new parameter \(t\) into the equation. As it turns out, the choice
\[I(t) = \int_0^{\infty} \frac{\tan^{-1}(tx)}{x(1+x^2)} \; dx\]
works well. When \(t = 1\),
\[I(1) = \int_0^{\infty} \frac{\tan^{-1}(x)}{x(1+x^2)} \; dx.\]
Also,
\[\begin{align} I(0) &= \int_0^{\infty} \frac{\tan^{-1}(0)}{x(1+x^2)} \; dx \\ &= \int_0^\infty 0 \; dx \\ &= 0. \end{align}\]
Your next step is to find \(I'(t)\):
\[\begin{align}I'(t) &= \frac{d}{dt}\int_0^{\infty} \frac{\tan^{-1}(tx)}{x(1+x^2)} \; dx\\&= \int_0^{\infty} \frac{\partial}{\partial t}\frac{\tan^{-1}(tx)}{x(1+x^2)} \; dx\\&= \int_0^{\infty} \frac{x}{(1+t^2x^2)x(1+x^2)} \; dx\\&= \int_0^{\infty} \frac{1}{(1+t^2x^2)(1+x^2)} \; dx. \end{align}\]
To continue the integration, use Integration by Partial Fractions:
\[ \begin{align} \int_0^{\infty} \frac{1}{(1+t^2x^2)(1+x^2)} \; dx &= \int_0^{\infty} \frac{-t^2/(1-t^2)}{1 + t^2x^2} \; dx + \int_0^{\infty} \frac{1/(1-t^2)}{1+x^2} \; dx \\&= -\frac{t}{1-t^2}\int_0^{\infty} \frac{t}{1 + t^2x^2} \; dx + \frac{1}{1-t^2}\int_0^{\infty} \frac{1}{1+x^2}\\&= -\frac{t}{1-t^2}\tan^{-1}(tx) + \frac{1}{1-t^2}\tan^{-1}(x)\bigg|_{x=0}^{x=\infty}. \end{align} \]
Now to evaluate the last term, use the fact that the arctangent function approaches \(\frac{\pi}{2}\) as \(x\) approaches infinity, so:
\[ \begin{align} I'(t) &= -\frac{t\pi}{2(1-t^2)} + \frac{\pi}{2(1-t^2)} + \frac{t}{1-t^2}(0) - \frac{1}{1-t^2}(0)\\&= \frac{\pi (t-1)}{2(t^2 - 1)}\\&= \frac{\pi}{2(t+1)}.\end{align}\]
Finally, use the Fundamental Theorem of Calculus to find \(I(1)\):
\[\begin{align}I(1) &= I(1) - I(0)\\&= \int_0^1 I'(t) dt\\&= \int_0^1 \frac{\pi}{2(t+1)} dt\\&= \frac{\pi}{2}\int_0^1\frac{1}{t+1} dt\\&= \frac{\pi}{2}\ln|t+1|\bigg|_{t=0}^{t=1}\\&= \frac{\pi}{2}\left[\ln|2| - \ln|1|\right]\\&= \frac{\pi}{2}\ln(2).\end{align}\]
As in this example, you often need to use Feynman integration with other techniques of integration.
Richard Feynman (1918-1988) was an American theoretical physicist who did significant work in particle physics and quantum mechanics. He was a brilliant physicist with a gift for explaining difficult concepts clearly, elegantly, and concretely. While he did not originate the integration technique that bears his name, he did play a role in popularizing it. Here is what he had to say on the technique:
“I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. One day he told me to stay after class. "Feynman," he said, "you talk too much and you make too much noise. I know why. You're bored. So I'm going to give you a book. You go up there in the back, in the corner, and study this book, and when you know everything that's in this book, you can talk again." ... [That book] showed how to differentiate parameters under the integral sign — it’s a certain operation. It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. So because I was self-taught using that book, I had peculiar methods of doing integrals. ... So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.”1
Techniques of Integration - Key takeaways
- Common techniques of integration include the Power Rule for Integrals, Integration by Substitution, Trigonometric Substitution, Integration by Parts, Integration by Partial Fractions, and Integrating Functions Using Long Division.
- The Power Rule for Integration is a rule that 'undoes' the power rule for differentiation.
- Weierstrass substitution is a useful substitution for rational expressions of trigonometric functions.
- The integral of a function can be expressed in terms of its inverse.
- Feynman's technique of integration is a useful technique for complicated integrals that involves differentiating under the integral sign.
References
- Richard Feynman, Surely You're Joking, Mr. Feynman!, 1985.
Learn faster with the 2 flashcards about Techniques of Integration
Sign up for free to gain access to all our flashcards.
Frequently Asked Questions about Techniques of Integration
What are the basic techniques of integration?
The basic techniques of integration are the power rule for integrals, integration by substitution, trigonometric substitution, integration by parts, integration by partial fractions, and integrating functions using long division.
What is the purpose of using integration techniques?
The purpose of using integration techniques is to simplify complicated integrals so that they can be more easily solved.
When to use techniques of integration?
Different integrals require different techniques of integration. With time and practice, it becomes easier to figure out which technique of integration ought to be used for which kinds of integral.
How do you use integration techniques?
How to use an integration technique depends on the integration technique you are using and the integral you are working with. The first step is to figure out which integration technique applies to the integral you are working with.
What are the different types of integration?
The two main types of integration are definite and indefinite integration. Definite integrals have bounds of integration, while indefinite integrals do not. There are many integration techniques that can be applied to solving each of these types of integrals.
About StudySmarter
StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.
Learn more