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Techniques of Integration

Techniques of Integration

Before getting into the math, here is a joke to lighten the mood (who said math couldn't be funny?):


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:

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

  • Trigonometric 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.)

Given a function \(f(x)\) that has inverse \(f^{-1}(x)\) and antiderivative,

\[\int f^{-1}(y) dy = yf^{-1}(y) - F(f^{-1}(y)) + C.\]

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:

Techniques of Integration Inverse Functions graph showing a function and the area that represents the integral of the function StudySmarterIntegral of a function as an area

You might remember that reflecting the graph of \(f\) about the line \(y=x\) gives the graph of \(f^{-1}\):

Techniques of Integration Inverse Functions graph showing a function reflected across the line y equals x to get the inverse of the function StudySmarterSymmetry 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.

Techniques of Integration Inverse Functions showing an integral as an area and the complementary area that is the inverse function integral  StudySmarterVisual 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.

Techniques of Integration Inverse Functions Visual proof of the integral of an inverse function with a non-zero limit of integration StudySmarterVisual 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.

Techniques of Integration triangle with angle and sides labeled to find arccosine StudySmarterTriangle 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

\[\int_a^b f(x) dx.\]

  1. 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 \[I(t) = \int_a^b x^t dx= \int_a^b f(x, t) dx.\]

  2. Find \(I'(t)\) by differentiating with respect to \(t\) under the integral sign.

    • In symbols, \[ I'(t) = \frac{d}{dt}\int_a^b f(x, t) dx = \int_a^b \frac{\partial}{\partial t} f(x, t) dx.\]

  3. 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

  1. Richard Feynman, Surely You're Joking, Mr. Feynman!, 1985.

Frequently Asked Questions about 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.

The purpose of using integration techniques is to simplify complicated integrals so that they can be more easily solved.

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 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.

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.

Final Techniques of Integration Quiz

Question

Integration by parts is an integration method based on:

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Answer

The Product Rule.

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Question

Which of the following is the integration by parts formula?

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Answer

\[ \int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u.\]

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Question

What does the “ I”  in LIATE stands for?

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Answer

Inverse Trigonometric Functions

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Question

Which trig substitution should you try for this integral?

\[\int\dfrac{\sqrt{x^2-25}}{x}\mathrm{d}x\]

Show answer

Answer

\[x=5\sec(\theta)\]

Show question

Question

Which trig substitution should you try for this integral?

\[\int\dfrac{x}{\sqrt{25-x^2}}\mathrm{d}x\]

Show answer

Answer

\[x=\tan(5\theta)\]

Show question

Question

Which trig substitution should you try for this integral?

\[\int\dfrac{x}{\sqrt{25+x^2}}\mathrm{d}x\]

Show answer

Answer

\[x=5\sin(\theta)\]

Show question

Question

Which of the following is the result of substituting \(x = 5\sin(\theta)\) in this integral:

\[\int\dfrac{\sqrt{x^2-25}}{x}\mathrm{d}x\]

Show answer

Answer

\[\int\dfrac{\sqrt{\sin^2(\theta)-25}}{\sin(\theta)}\mathrm{d}\theta\]

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Question

What conditions must the function \(g\) satisfy to be able to make the inverse substitution \(x=g(\theta)\)?

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Answer

\(g\) must be one-to-one and differentiable.

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Question

What is the Inverse Substitution Rule?

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Answer

The Inverse Substitution Rule states that, given an integrable function \(f\) and a differentiable, one-to-one function \(g\), 


\[\int f(x) \mathrm{d}x=\left.\int f(g(\theta))g'(\theta) \mathrm{d}\theta\right|_{\theta=g^{-1}(x)}\]

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Question

Given that \(x = 5\sin(\theta)\), what is the value of \(\tan(\theta)\)?

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Answer

\[\tan(\theta)=\dfrac{x}{\sqrt{25-x^2}}\]

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Question

What integral is the result of substituting \(x = 5\tan(\theta)\) in the integral:

\[\int \dfrac{x}{\sqrt{x^2+25}} \mathrm{d}x\]

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Answer

\[\int \dfrac{5\tan(\theta)}{\sqrt{25\tan^2(\theta)+25}}5\sec^2(\theta) \mathrm{d}\theta\]

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Question

Why does trigonometric substitution work even though many trigonometric functions are not one-to-one?

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Answer

Trigonometric substitution still works since we can restrict the values of theta to ranges over which the trigonometric function we are using is one-to-one.

Show question

Question

What is the result of substituting \(x=5\sec(\theta)\) in the integral:

\[\int\dfrac{\sqrt{x^2-25}}{x}\mathrm{d}x\]

Show answer

Answer

\[\int\dfrac{\sqrt{25 \sec^2(\theta)-25}}{5 \sec(\theta)}5 \sec(\theta) \tan(\theta)\mathrm{d}\theta\]

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Question

If you make the substitution u = g(x), what is du in terms of dx?

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Answer

du = g'(x)dx

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Question

When should you use u-substitution?

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Answer

In general, u-substitution is a good strategy to try whenever you have an integrand that is a composite function (of the form f(g(x)) for some functions f and g).

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Question

What differentiation rule does integration by substitution "undo"?

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Answer

The Chain Rule

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Question

What differentiation technique does the Power Rule for Integration 'undo'?

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Answer

The Power Rule for Differentiation

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Question

When should you use Weierstrass substitution?

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Answer

Weierstrass substitution applies to integrals that are rational functions of sine and cosine

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Question

What does this symbol mean: \( \frac{\partial}{\partial t} \)?


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Answer

This symbol means 'differentiate with respect to \(t\), not with respect to any other variable'

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Question

What is another term for Feynman's Technique of Integration?

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Answer

Differentiating under the integral sign

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Question

When defining \(I(t)\) for Feynman integration, what conditions should you make sure \(I(t)\) satisfies?

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Answer

\(I(0) = 0\) and there exists some a such that \(I(a) =\) your integral

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Question

Which of the following is the solution of:

\[ \int x^3 \; dx \]


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Answer

\( 3x^2 \)

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Question

The power rule for integration can be used to solve 

\[ \int x^{-\pi} \; dx \]


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Answer

True

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Question

Find

\[ \int x^{-3} \; dx \]


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Answer

\[ -\frac{1}{2}x^{-2} + C \]

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Question

In a few words, what is Partial Fraction Decomposition?

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Answer

Partial Fraction Decomposition is the process of rewriting a rational expression as a sum or difference of simpler fractions.

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Question

What is the first step for adding two fractions?

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Answer

The first step to add two fractions is finding a common denominator.

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Question

Describe Integration by Partial Fractions in a few words.

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Answer

Integration by Partial Fractions consists in doing the partial fraction decomposition of an integral, so it can be easier to evaluate.

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Question

Suppose that in a partial fraction decomposition there is a quadratic factor with a discriminant equal to 5. Which of the following is true?

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Answer

The quadratic factor can be split into two different factors.

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Question

Suppose that in a partial fraction decomposition there is a quadratic factor with a discriminant equal to 0. Which of the following is true?

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Answer

The quadratic factor can be split into two repeated factors.

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Question

Suppose that in a partial fraction decomposition there is a quadratic factor with a discriminant equal to -8. Which of the following is true?

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Answer

The quadratic factor cannot be reduced.

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Question

Consider the integral \[ \int x^2 e^x\, \mathrm{d}x.\] According to LIATE, which should be your choice of \(u\)?

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Answer

\( u=x^2.\)

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Question

Consider the integral
\[ \int x^3\sin{x}\,\mathrm{d}x. \]
According to LIATE, which should be your choice of \(u\)?

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Answer

\( u=x^3.\)

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Question

Consider the integral
\[ \int x^2\ln{x}\,\mathrm{d}x. \]
According to LIATE, which should be your choice of \( u\)?

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Answer

\( u=\ln{x}.\)

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Question

Consider the integral
\[  \int e^x\sin{x}\,\mathrm{d}x.\]
According to LIATE, which should be your choice of \(u?\).

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Answer

\( u=\sin{x}.\)

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Question

Consider the following integral
\[\int \arctan{x}\,\mathrm{d}x.\]
Can you use Integration by Parts on this integral?

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Answer

Yes, let \( u=\arctan{x}\) and \( \mathrm{d}v=\mathrm{d}x.\)

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Question

In order to solve \[\int x^2 \sin{x}\,\mathrm{d}x \] the substitutions \( u=x^2\) and \( \mathrm{d}v = \sin{x}\) have incorrectly been made. What is the mistake?

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Answer

\( \sin{x} \) is missing \( \mathrm{d}x.\)

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Question

How do you solve a definite integral using Integration by Parts?

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Answer

Find the indefinite integral first by using Integration by Parts, then, evaluate it using the Fundamental Theorem of Calculus.

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Question

When do you use integration by long division?

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Answer

When you have \( \int \frac{p(x)}{q(x)} \; dx,\) where \( p(x)\) and \(q(x)\) are both polynomials and \( \deg⁡(p(x))≥\deg⁡(q(x))\).


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Question

After doing the long division, what would the integrand look like?

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Answer

\[\int \frac{p(x)}{q(x)} \; dx = \int s(x) +\frac{r(x)}{q(x)} \; dx\]

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Question

To solve the following integral you must use long division \[ \int \frac{x^3 - 3x^2 + 2}{x^4 - 1} \; dx\]

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Answer

False

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Question

What is the difference between integration by long division and integration by parts?

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Answer

In long division, the numerator has a greater degree than the denominator, while integration by parts the numerator has a smaller degree than the denominator 

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Question

What are the steps to apply long division integration?

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Answer

1 - Verify that the numerator's degree is bigger or equal to the denominator.

2 - Use long division to write \( \frac{p(x)}{q(x)} =  s(x) + \frac{r(x)}{q(x)} \).

3 - Integrate following the integration rules.


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Question

Rewrite the following integral using long division \[ \int \frac{x^3 - 3}{x - 2} dx\]

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Answer

\[ \int \frac{x^3 - 3}{x - 2} dx = \int x^2 + 2x + 4 + \frac{5}{x-2} dx\]

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Question

To solve the following integral you must use long division \[ \int \frac{x^4 - 3x^2 + 2}{x - 2x^2-x^4 } \; dx\]

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Answer

True

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Question

If you have \( \int \frac{p(x)}{q(x)} \; dx,\) where \( \deg⁡(p(x))=\deg⁡(q(x))\), then you cannot apply long division.

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Answer

False

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Question

What is the degree of \[p(x)=x^7-7x^5+x-2?\]

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Answer

\[\deg(x^7-7x^5+x-2 ) = 7\]

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Question

The degree of a polynomial is the exponent of the highest power.

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Answer

True

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Question

After applying the long division technique, if you have \( r(x)=0 \) then you will be integrating a polynomial.

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Answer

True

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Question

What is the most appropriate technique to solve the following integral? \[\int \dfrac{1}{\sqrt{x^2+2x+2}}\mathrm{d}x\]

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Answer

First - Complete the square inside the square root.

Second - Do a \(u\)-substitution.

Third - Do a trigonometric substitution.

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Question

What is the most effective substitution for \( x^2+a^2\)?

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Answer

\(x=a\tan(\theta)\)

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