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You CAN Ace Calculus

17calculus > integrals > integration by substitution

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Integration by Substitution

Integration by substitution is sometimes called u-substitution since u is often used as the substitution variable.

Integration by substitution is the first major integration technique that you will probably learn and it is the one you will use most of the time. In fact, as you learn more advanced techniques, you will still probably use this one also, in addition to the more advanced techniques, even on the same problem. So, this is a critically important technique to learn.

The main idea of this technique is to reduce the complexity of an integral by introducing a new variable to replace a part of the integrand. Then we convert the entire integrand in terms of the new variable. This should reduce the integral to one that we can evaluate using another technique.

 Where Integration by Substitution Comes From

The idea of integration of substitution comes from something you already now, the chain rule. Remember, the chain rule for $$y=f(g(x))$$ looks like $$y'=f'(g(x))g'(x)$$. Integration by substitution reverses this by first giving you $$f'(g(x))g'(x)$$ and expecting you to come up with $$f(g(x))$$. This is easier than you might think and it becomes easier as you get some experience. The key is to know how to choose $$g(x)$$. First, let's see how the chain rule becomes an integral.

As we said above, the chain rule for $$y=f(g(x))$$ is $$y'=f'(g(x))g'(x)$$. This can also be written $$dy/dx=f'(g(x))g'(x)$$ or in differential form $$dy=f'(g(x))g'(x)~dx$$. Now when we integrate both sides we have $$y=\int{ f'(g(x))g'(x)dx }$$. The key is to let $$u=g(x)$$ and then $$du=g'(x)dx$$. The integral can now be written $$y=\int{ f'(u)~du } = f(u)+C = f(g(x))+C$$.

 How To Choose u

Let's watch a video with a VERY quick example. In this video she gives some really good advice on what to choose for u.

 Krista King Math - Solving Integrals

Whew! That video clip went through the example pretty quickly without much explanation. Let's take it a bit slower and explain what is going on. In the video, the presenter evaluated the integral $$\int{x \sqrt{x^2+1} ~dx}$$. Using her ideas, she chose $$u=x^2+1$$. She actually lays out the solution very well but doesn't allow you much time to digest it. So, go back to the video and pause it at the moment she shows her complete solution and make sure you can follow each step.

There are some specific guidelines to get you started as far as what to choose for u. Let's go through them and show you some examples.

Guideline 1 - Term Under A Power or Root

The video above gives one possible choice for u, the higher power when you have another term with a power one less. From this example, I would extend that idea a bit to include choosing u as whatever is under a radical or a term with a power (like $$(x^2+3)^3 \to u=x^2+3$$).

Let's look at an example to see how to apply this idea.

 Evaluate $$\int{ 3x^2 \sqrt{x^3+1} ~dx}$$.

Notice in the solution to the last example, that at one point we had x's and u's in the integral. We could not evaluate the integral until it had only the one variable u. So had to cancel the $$3x^2$$ term in the numerator and denominator before we could actually integrate.

Okay, now try some on your own. This is one technique that you want to practice a lot until you can do it in your sleep since you will be using in almost every integral that you evaluate.

 Basic Problems

Practice 1

$$\displaystyle{\int{(3x-5)^{17}dx}}$$

solution

Practice 2

$$\displaystyle{\int{x\sqrt{x^2+9}~dx}}$$

solution

Practice 3

$$\displaystyle{\int{x^2\sqrt{2x^3-4}~dx}}$$

solution

Practice 4

$$\displaystyle{\int{2x(x^2+1)^{30}~dx}}$$

solution

Practice 5

$$\displaystyle{\int{2x(x^2+4)^{100}~dx}}$$

solution

Practice 6

$$\displaystyle{\int{y\sqrt[5]{3y^2+1}~dy}}$$

solution

Practice 7

$$\displaystyle{\int{(12x+8)(3x^2+4x+1)^9~dx}}$$

solution

 Intermediate Problems

Practice 8

$$\displaystyle{\int{\frac{1}{(4x+7)^6}dx}}$$

solution

Practice 9

$$\displaystyle{\int{\frac{4}{(3x+1)^5}~dx}}$$

solution

Practice 10

$$\displaystyle{\int{\frac{4x}{\sqrt{x^2+1}}dx}}$$

solution

Practice 11

$$\displaystyle{\int{\frac{(\ln(x))^{10}}{x}dx}}$$

solution

Practice 12

$$\displaystyle{\int{ \csc x \cot x \sqrt{1-\csc x}~dx}}$$

solution

Guideline 2 - Exponential

In the case of an exponential, choose u as the exponent. Since we know how to evaluate $$\int{e^u~du}$$, this choice should significantly simplify the integral.

 Basic Problems

Practice 13

$$\displaystyle{\int{e^{-x}~dx}}$$

solution

Practice 14

$$\displaystyle{\int{\frac{e^x+1}{e^x}dx}}$$

solution

Practice 15

$$\displaystyle{\int{\frac{e^{\sqrt{x}}}{\sqrt{x}}dx}}$$

solution

Practice 16

$$\displaystyle{ \int{ \frac{e^{1/x}}{x^2}~dx } }$$

solution

 Intermediate Problems

Practice 17

$$\displaystyle{ \int{ (x^2+x)e^{2x^3+3x^2}~dx }}$$

solution

Practice 18

$$\displaystyle{\int{e^{x+e^x}~dx}}$$

solution

Guideline 3 - Trig Functions

When sine and cosine functions are involved, choose u as one of them. When you take the differential, the other one will be part of the differential. For example, $$u=sin(x) \to du=cos(x)~dx$$. Also, it sometimes helps to choose u as the angle of a trig function, if the angle is complicated. For example, if you have $$\sin(3t)$$, let $$u=3t$$.

 Basic Problems

Practice 19

$$\displaystyle{ \int{(\sin x) e^{\cos x}~dx}}$$

solution

Practice 20

$$\displaystyle{\int{\tanh(\omega t)~dt}}$$

solution

 Intermediate Problems

Practice 21

$$\displaystyle{ \int{ \left[ \sec(\cos x)\tan(\cos x )\right]\sin x ~dx}}$$

solution

Guideline 4 - Fraction Denominator

When it makes sense, sometimes choosing u as the denominator of a fraction is helpful since $$\int{(1/x)dx} = \ln(x)$$. Let's look at an example.

 Evaluate $$\displaystyle{ \int{ \frac{x^3}{5+x^4} ~dx} }$$.

In the last example, notice that all the x terms canceled but we had an extra constant, which we factored out of the integral. This will happen quite often and it helps to move all the constants outside the integral like we did so that it is easy to see that all the variables in the integral are now u's. Once we have that, we can do the actual integration.

 Basic Problems

Practice 22

$$\displaystyle{\int{\frac{5x}{5+2x^2}dx}}$$

solution

Practice 23

$$\displaystyle{\int{\frac{2^x}{2^x+1}dx}}$$

solution

Practice 24

$$\displaystyle{\int{ \frac{x}{x^2+4}~dx } }$$

solution

Practice 25

$$\displaystyle{\int{ \frac{24x^3-4}{3x^4-2x+1}~dx } }$$

solution

 Intermediate Problems

Practice 26

$$\displaystyle{ \int{\frac{1}{x^2+4x+13}dx} }$$

solution

Practice 27

$$\displaystyle{\int{\frac{dx}{1+x^{1/4}}}}$$

solution

Practice 28

$$\displaystyle{\int{\frac{x}{1-x^2+\sqrt{1-x^2}}dx}}$$

solution

Practice 29

$$\displaystyle{\int{\frac{x}{x^2+6x+10}~dx}}$$

solution

 Extra x Terms

Up until now, we have done the substitution of u and all the x terms have canceled. However, what do you do when they do not all cancel? Let's look at an example to see how to do this.

 Evaluate $$\int{ x \sqrt{x+2} ~dx }$$.

Here are some practice problems involving this technique.

Practice 30

$$\displaystyle{\int{\frac{x}{\sqrt[4]{x+2}}dx}}$$

solution

Practice 31

$$\displaystyle{\int{x\sqrt[3]{x+4}~dx}}$$

solution

Practice 32

$$\displaystyle{ \int{ x\sqrt{2x+3}~dx } }$$

solution

 Integration By Substitution and Definite Integrals

Note: This section requires understanding how to evaluate definite integrals.

When you have a definite integral, you need to do additional work to handle the limits of integration. Additionally, there is a mistake that many students make that could not only cause you to get an incorrect answer but also, if your instructor is paying attention, could cost you extra points. First, let's talk about the correct way to handle the limits of integration and then show you how NOT to do it.

There are basically two techniques, both of them correct but pay attention to what your instructor requires, since they may want you to use one of them and not the other. The first and best (in our opinion) way is to use the expression for u to change the limits of integration. Let's do an example to see how this works.

 Evaluate $$\displaystyle{ \int_{-2}^{1}{ x \sqrt{x+2} ~dx } }$$.

Notice in that example, we changed the limits of integration from what they were in terms of x to their values when the variable to integration is changed to u. The limits of integration are tied to the variable of integration. So we could write the integral as $$\displaystyle{ \int_{x=-2}^{x=1}{ x \sqrt{x+2} ~dx } }$$

Okay, so that is the first way and, in our opinion, the best way to handle definite integrals when we use integration by substitution. The other way is to drop the limits of integration, evaluate the integral, go back to x's in the integral and then substitute the original limits of integration. This next example, shows how to do that.

 Evaluate the same integral as the last example, but do not change the limits of integration and use correct notation.

Notice in the solution to the last example, we dropped the limits of integration when we wrote the integral in terms of u. It would be incorrect to write $$\displaystyle{ \int_{-2}^{1}{ (u-2)\sqrt{u} ~du} }$$ since the limits of integration are in terms of x but the integral is in terms of u. Now, theoretically it MIGHT be okay to write $$\displaystyle{ \int_{x=-2}^{x=1}{ (u-2)\sqrt{u} ~du } }$$ but no one does that (but, of course, check with your instructor to see what they expect). So not only is the notation $$\displaystyle{ \int_{-2}^{1}{ (u-2)\sqrt{u} ~du } }$$ incorrect, but after you have completed the integration it is extremely easy to forget to go back to x's in the integral before substituting the limits of integration.

As an instructor, I ALWAYS take off points for writing the integral this way. So be extra careful. We recommend that you always change the limits of integration like we did in the first example. Here is a video that explains very well when to change the limits of integration, why instructors are so picky about it and possible mistakes that could trip you up. We highly recommend this video.

 PatrickJMT - U-Substitution: When Do I Have to Change the Limits of Integration?

Okay, time for some practice problems.

Practice 33

$$\displaystyle{\int_{1}^{2}{\frac{2x}{\sqrt{x^2+1}}~dx}}$$

solution

Practice 34

Show that $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = \int_{a}^{b}{f(a+b-x)~dx} }$$

solution

Practice 35

$$\displaystyle{\int_{0}^{4}{x\sqrt{x^2+9}~dx}}$$

solution

Practice 36

$$\displaystyle{ \int_2^5{ x^2\sqrt{x^3-4}~dx } }$$

solution

Practice 37

$$\displaystyle{ \int_{1}^{e^3}{ \frac{\ln\sqrt{x}}{x}~dx } }$$

solution