# Section 11: Methods of Integration

## 11.5 Trigonometric Substitution

In the days before telephone service became widespread, a gentleman named Mr. Nakamura emigrated from Japan to San Francisco, bringing his aging mother along with him. They shared an apartment together, and Mr. Nakamura, who was an adept businessman, soon prospered in his new home. It wasn't long before he was expanding his business to take advantage of opportunities in Los Angeles as well. But to secure those opportunities Mr. Nakamura had to make a trip to Los Angeles, leaving his poor mother home alone. The thought of his mother fending for herself in a land that still seemed alien to her troubled Mr. Nakamura more and more each day he was away. On the fifth day he decided he needed to send his mother a message of reassurance that he would be returning soon. So he went to the Western Union office to send her a telegram.

Mr. Nakamura had written his message to his mother out in Japanese characters. But the telegraph operator had no idea how to encode Japanese characters into dots and dashes. So he suggested to Mr. Nakamura that he translate the message into English.

"Oh no," said Mr. Nakamura. "That would never do. My mother reads only Japanese such as you see here. A message in English might as well be in Persian as far as she's concerned."

"We could have somebody translate it back into Japanese at the other end," suggested the Western Union man.

Mr. Nakamura thought for a moment, then said, "No that won't do either. The words wouldn't be the same after two translations. My mother needs to read my exact words, for I have written my message to her in Haiku, and only the exact words will preserve the poetic form."

Mr. Nakamura stood silent for several minutes thinking. Then he said, "Instead, I shall spell out each of the Japanese words in my message phonetically, using your alphabet. You send that over the wire. Your Japanese translator at the other end will certainly recognize the words by their phonetic spellings, so he shall be able to transcribe them back into Japanese characters that my mother can read. Yes, and she will read my exact words and know how deeply her only son is missing her."

Imagine that the methods you've learned so far for finding the integral of a function are like passing a message over the telegraph wires. Now imagine that you have an integral on which none of the tools you've learned so far work. It is like Mr. Nakamura's message. It is spelled in an alphabet that renders your methods useless. The solution to the problem could very well be to transcribe your function into an alphabet that your methods do work on. You could then apply your method, then translate it back into its original alphabet when you're done.

Here is an example of doing just that:

 ``` ``` ``` dx ______ eq. 11.5-1 √1 - x2 ```
Your first instinct might be to substitute  u = 1 - x2.  But if you tried that you would soon find that it didn't work. When you took the derivative of u, you'd find  du = -2x dx,  and there is no place to make that substitution in this integral.

In the section on simple substitution you learned how to make substitutions based upon finding some function,  u(x),  that you could slip in to simplify your integral. What you haven't tried yet is turning this method on its head. That is, we imagine that our independent variable, x, is actually a function of the substitution variable, u. Suppose we tried that substitution blindly in the example above. Then we'd have

```  x  =  f(u)                                                      eq. 11.5-2a
```
for some mysterious and as yet unknown function, f. Applying implicit differentiation to this we'd also have
```               du
1  =  f'(u)                                                    eq. 11.5-2b
dx
```
or more appropriately for the purpose of substitution
```  dx  =  f'(u) du                                                 eq. 11.5-2c
```
Now we put all that into the integral, substituting  f(u)  for x and  f'(u) du  for dx:

 ``` ``` ``` f'(u) du _________ eq. 11.5-3 √1 - f2(u) ```
If only we could find a function,  f(u),  such that

```              _________
f'(u)  =  √1 - f2(u)                                           eq. 11.5-4
```
it would make life wonderful, because you could then substitute the right-hand side of equation 11.5-4 for  f'(u)  into the integral in 11.5-3, and all the hard stuff would cancel. You'd be left having only to integrate

 ``` ``` ``` du eq. 11.5-5 ```
which is a piece of cake. Well I just happen to know of such a function. Remember from trigonometry that

```   sin2(u) + cos2(u)  =  1                                        eq. 11.5-6a
```
or equivalently
```               ___________
cos(u)  =  √1 - sin2(u)                                        eq. 11.5-6b
```
But we already know that cosine is the derivative of the sine. So the function,  f(u) = sin(u)exactly satisfies our fondest desires for  f(u)  as expressed in equation 11.5-4. Now let's do the whole substitution exercise over again using
 ``` x = sin(u) ``` and ``` ___________ dx = cos(u) du = √1 - sin2(u) du ``` ``` eq. 11.5-7 ```

So using those substitutions for x and dx, the integral becomes

 ``` ``` ``` dx ______ = √1 - x2 ``` ``` ``` ``` ___________ √1 - sin2(u) du ___________ = √1 - sin2(u) ``` ``` ``` ``` du = u + C eq. 11.5-8 ```
At this point we have, in a manner of speaking, transcribed the Japanese message into our English alphabet, sent it over the wire, and all we have left to do is transcribe it back to Japanese characters so that Mr. Nakamura's mother can read it. The original substitution was  x = sin(u).  So what do we put in for u? We have to use inverse sine or

```   arcsin(x)  =  u                                                eq. 11.5-9
```
and we find that when we substitute back we get

 ``` ``` ``` dx ______ = arcsin(x) + C eq. 11.5-10 √1 - x2 ```

You might have guessed this integral from our table of common derivatives. You will see the relationship between arcsin and our integrand near the bottom of that table. All we did was a fancy parlor trick to solve an integral that we already knew. But this same attack can integrate functions that are not so obvious. For example, let's integrate

 ``` ``` ``` ______ √1 - x2 dx eq. 11.5-11 ```

We'll try the same substitution as before of
 ``` x = sin(u) ``` and ``` ___________ dx = cos(u) du = √1 - sin2(u) du ```

 ``` ``` ``` ______ √1 - x2 dx = ``` ``` ``` ``` ___________ √1 - sin2(u) ``` ``` ___________ √1 - sin2(u) du = ```
 ``` ``` ``` (1 - sin2(u)) du = ``` ``` ``` ``` cos2(u) du eq. 11.5-12 ```

And now you're left having to integrate the square of the cosine. But this one isn't so bad once you apply a trig identity to it (which illustrates why you should sharpen up your command of trig identities for this unit. You will need them time and again). The integral becomes a sum that you can break up:

 ``` ``` ``` cos2(u) du = ``` ``` 1   2 ``` ``` ``` ``` (1 + cos(2u)) du = ```
 ``` 1   2 ``` ``` ``` ``` ``` ``` du + ``` ``` ``` ``` cos(2u) du ``` ``` ``` ``` eq. 11.5-13a ```

Both of these are integrals you know how to do (and if you need to you can still use the table and equation 11.2-9).

 ``` 1   2 ``` ``` ``` ``` ``` ``` du + ``` ``` ``` ``` cos(2u) du ``` ``` ``` ``` 1 1 = u + sin(2u) + C eq. 11.5-13b 2 4 ```

To translate back to our original variable, x, you will have to use another trig identity.

 ``` 1   2 ``` ``` ``` ``` ``` ``` du + ``` ``` ``` ``` cos(2u) du ``` ``` ``` ``` 1 1 = u + sin(u)cos(u) + C eq. 11.5-13c 2 2 ```

Now we can substitute back. The original substitution was  sin(u) = x.  So running that in reverse on the bare u in equation 11.5-13c, it becomes  arcsin(x).  The  sin(u)  becomes x. And we can use this to put  cos(u)  in terms of  sin(u)  to get an x expression for it as well. Hence

 ``` ``` ``` ______ √1 - x2 dx = ``` ``` 1 1 u + sin(u)cos(u) + C = 2 2 ```
 ``` 1 arcsin(x) + 2 ``` ``` 1 ______ x √1 - x2 + C eq. 11.5-13d 2 ```

I'd like you to take the time right now to take the derivative of this result so that you can see how intimately the terms above conspire to give you back the original integrand (which is a nice way of saying that you'll have to do some algebraic manipulations on the immediate derivative to get all the way back to the original integrand).

As nicely as that works out, we have applied this method only to a very narrow collection of integrands. Supposing you have a variation on the theme. How would you integrate

 ``` ``` ``` dx ______ eq. 11.5-14a √9 - x2 ```

The plan here is to do a little algebraic munging to turn this into something that we can apply a substitution similar to the one we used on the first example. First, factor 9 out of the expression under the radical. Remember that when you pull that factor out of the radical, it gets square-rooted.

 ``` 1   3 ``` ``` ``` ``` dx __________ eq. 11.5-14b √1 - (x/3)2 ```

The substitution is almost the same, but we have to account for the factor of 1/3 that now goes with the x

Alternative method to integrate

 ``` ``` ``` dx ______ √9 - x2 ```

Skip factoring out the 1/3.
Since  3 = √9,  substitute

```   3 sin(u)  =  x
```
and you still get
```   3 cos(u) du  =  dx
```
Since  3 cos(u) = √9 - x²,  you will find that you end up with the same expression to integrate as if you had done it the way the main text suggests. Try it (it's really the same method).

 ``` x sin(u) =   3 ``` and ``` dx cos(u) du =   3 ```

And that means that  3 cos(u) du = dx.  I'll let you finish this one off. In the end make sure you get

 ``` ``` ``` dx ______ = arcsin(x/3) + C √9 - x2 ```

```                               eq 11.5-15
```

Let's try to apply this method to an integral that has an additional complication.

 ``` ``` ``` dx ___________ eq. 11.5-16 √3 - 2x - x2 ```

When you see a quadratic under the radical, the first step always is to complete the square. If there is no coefficient on the x2 term and the x2 is negative, then call the quadratic you have under the radical,  -q(x).  That makes  q(x) = x2 + 2x - 3.  simply take half the middle coefficient of  q(x)  and add it to x. That will be your first substitution. The middle coefficient for this  q(x) is 2, hence

 ``` v = x + 1 ``` and ```dv = dx ``` ``` eq. 11.17a ```

But look what happens when you square v

```   v2  =  x2 + 2x + 1                                              eq. 11.5-17b
```
and then add or subtract whatever it takes from both sides to make the right-hand side turn into the negative of what's under the radical:
```   v2 - 4  =  x2 + 2x - 3
eq. 11.5-17c
4 - v2  =  3 - 2x - x2
```
With this substitution, the integrand is now in a form we know how to integrate:

 ``` ``` ``` dx ___________ = √3 - 2x - x2 ``` ``` ``` ``` dv ______ = √4 - v2 ``` ``` 1   2 ``` ``` ``` ``` dv __________ √1 - (v/2)2 eq. 11.5-18a ```

Now it is just like the last one we did. You can substitute

 ``` v sin(u) =   2 ``` and ``` dv cos(u) du = eq. 11.5-18b 2 ```

to find that

 ``` ``` ``` dv ______ = √4 - v2 ``` ``` arcsin(v/2) + C eq. 11.5-18c ```

But you have one more step to do, and that is substitute back to an expression in x. Remember that the first substitution we made was  v = x + 1.  Wherever there's a v, replace it with  x + 1

 ``` ``` ``` dx ___________ = √3 - 2x - x2 ``` ``` ``` ``` dv ______ = √4 - v2 ```
 ``` arcsin(v/2) + C = arcsin((x+1)/2) + C eq. 11.5-19 ```

### Exercises

1) Use trigonometric substitution to find the indefinite integral:

 ``` ``` ``` _______ √25 - x2 dx ```

2) Use completing the square followed by trigonometric substitution to find the indefinite integral:

 ``` ``` ``` ____________ √35 + 2x - x2 dx ```

Check back here if you need to review how to integrate the square of the cosine on either of the above two problems.

3) Use trigonometric substitution to find the indefinite integral:

 ``` ``` ``` dx _______ √1 - 9x2 ```

Scroll down for more on Trigonometric Substitution

# More Trig Substitution

Trigonometric substitution is a big topic because there are lots of variations. Only with practice will you gain the skill of knowing what variation to use on each integrand. Here is an integrand that you've seen before because it's on the table, but it is useful to see how we can use a trigonometric substitution on it.

 ``` ``` ``` dx eq. 11.5-20 1 + x2 ```

Here the temptation is to try simple substitution using  u = 1 + x2,  but you will find that again you have no place to substitute the  du = 2x dx  that arises as a result.

So let's ask ourselves, what would the ideal substitution function,  f(u) = x  have to do in order to be ideal for this integrand? Taking the derivative implicitly you have

 ``` du f'(u) = 1 dx ``` or ``` f'(u) du = dx eq. 11.5-21 ```

To make all the hard stuff cancel, we want an  f(u)  that has the following property:

```   f'(u)  =  1 + f2(u)                                            eq. 11.5-22
```
because then the substituted integral will become

 ``` ``` ``` dx = 1 + x2 ``` ``` ``` ``` (1 + f2(u)) du = 1 + f2(u) ``` ``` ``` ``` du eq. 11.5-23 ```

Once again, I just happen to know of such a function. Go to the table of common derivatives and look up the derivative of the tangent function. Can you see that if we make the substitution

 ``` tan(u) = x ``` and ``` (1 + tan2(u)) ``` ``` du = 1 dx ``` or ``` (1 + tan2(u)) du = dx ``` ``` eq. 11.5-24 ```

then all that's left in the substituted integral is

 ``` ``` ``` dx = 1 + x2 ``` ``` ``` ``` du = u + C eq. 11.5-25a ```

Then substituting back, we have to apply the inverse function of our original substitution in order to back-substitute the bare u.

 ``` ``` ``` dx = 1 + x2 ``` ``` arctan(x) + C eq. 11.5-25b ```

Once again we have gone to some trouble to integrate something we already knew the integral of. But what happens when you have a more complicated quadratic in the denominator? Suppose we had

 ``` ``` ``` dx eq. 11.5-26 40 + 12x + x2 ```

The first thing to do is to complete the square. Adding half the middle coefficient to x gives

 ``` v = x + 6 ``` and ``` v2 = x2 + 12x + 36 eq. 11.5-27a ```

Now what do we add to or subtract from v2 to make it be equal to the original  q(x)

```   v2        =  x2  +  12x  +  36
+  4  =              +   4

v2  +  4  =  x2  +  12x  +  40
```
Taking the derivative of  v = x + 6,  we find that  dv = dx

So the completed square integral is

 ``` ``` ``` dx = 40 + 12x + x2 ``` ``` ``` ``` dv eq. 11.5-28a 4 + v2 ```

Note that trig substitution using the tangent function will work only if the denominator has no real roots. That is, at this point you must have ended up with a denominator in the form of a positive quantity added to v2. If you end up with the difference between v2 and a positive quantity, this substitution won't work. We will deal with that case in a later section.

Now factor out the 4 from the denominator:

 ``` ``` ``` dv = 4 + v2 ``` ``` 1   4 ``` ``` ``` ``` dv eq. 11.5-28b 1 + (v/2)2 ```

Now we make our trig substitution:

 ``` v tan(u) =   2 ``` and ``` (1 + tan2(u)) ``` ``` du 1 =   dx 2 ``` or ``` 2(1 + tan2(u)) du = dx ```

Substituting and integrating:

 ``` 1   4 ``` ``` ``` ``` dv = 1 + (v/2)2 ``` ``` 1   2 ``` ``` ``` ``` 1 du = u + C eq 11.5-29 2 ```

Now substitute back to v using the inverse function

 ``` 1   4 ``` ``` ``` ``` dv = 1 + (v/2)2 ``` ``` 1 arctan 2 ``` ``` ``` ```v   2 ``` ``` ``` ``` + C eq. 11.5-30a ```

And finally substitute back to x using  v = x + 6

 ``` ``` ``` dx = 40 + 12x + x2 ``` ``` 1 arctan 2 ``` ``` ``` ```x + 6   2 ``` ``` ``` ``` + C eq. 11.5-30b ```

### A Broader Class of Integrands

The trig substitutions we have explored so far combined with completing the square can integrate

 ``` ``` ``` dx   b + ax + x2 ```

whenever the quadratic has no real roots (which is the case if and only if  a2 < 4b)  and

 ``` ``` ``` dx ___________ √b - ax - x2 ``` and ``` ``` ``` ___________ √b - ax - x2 dx ```

whenever the quadratic under the radical does have real roots (which is the case if and only if a2 + 4b ≥ 0).  The truth is that the method still addresses only a narrow class of functions. In this paragraph, we show how we can still use this method whenever the integrand is any of the functions listed above times some polynomial function.

Suppose we needed to integrate

 ``` ``` ``` 15 - 7x + 3x2 dx eq. 11.5-31 1 + x2 ```

The strategy is to do some algebraic munging to put this into a form that we do know how to integrate. And the algebraic tool that we use to do this is polynomial long division.

```          3
x2 + 1)3x2 - 7x + 15
3x2 + 0x +  3

-7x + 12
```
That is we get a quotient of 3 and a remainder of  -7x + 12 You need to remember the following very important rule concerning polynomial long division: The Quotient added to the Remainder over the Denominator is always equal to the Numerator over the Denominator.
 ``` R(x) N(x) Q(x) + = eq. 11.5-32 D(x) D(x) ```
In the case here, you have  Q(x) = 3,     R(x) = -7x + 12,     D(x) = x2 + 1,  and  N(x) = 3x2 - 7x + 15.  Hence we can rewrite the integral as

 ``` ``` ``` 15 - 7x + 3x2 dx = 1 + x2 ``` ``` ``` ``` ``` ``` -7x + 12 3 +   1 + x2 ``` ``` ``` ``` dx = ```
 ``` 3 ``` ``` ``` ``` dx - 7 ``` ``` ``` ``` x dx + 12 1 + x2 ``` ``` ``` ``` dx eq. 11.5-33 1 + x2 ```

All three of these integrals you know how to do. The first is trivial. On the second you apply simple substitution. And on the third you apply trig substitution (or you find it on the table). We find that

 ``` ``` ``` 15 - 7x + 3x2 dx = 1 + x2 ```

```                7
3x  -    ln|1 + x2|  +  12 arctan(x)  +  C               eq. 11.5-34
2
```

Let's try a more difficult one. Integrate

 ``` ``` ``` 3 - 2x - 14x2 + x3 eq. 11.5-35 85 + 18x + x2 ```

There are several ways you can attack this that will work. I prefer doing the polynomial long division first (you could complete the square first, but then you'd have to apply the substitution that arises from that to the cubic in the numerator).

Step 1: Do the long division. The long division is

```                         x  - 32
x2 + 18x + 85 ) x3 - 14x2 -   2x +    3
x3 + 18x2 +  85x

- 32x2 -  87x +    3
- 32x2 - 576x - 2720

489x + 2723
```
If the polynomial long division here confuses you, you should review that algebra topic until you are comfortable with it. With the quotient of  x - 32  and the remainder of  489x + 2723  shown here, and applying equation 11.5-32, you can break the original integral up into:

 ``` ``` ``` 3 - 2x - 14x2 + x3 = eq. 11.5-36 85 + 18x + x2 ```
 ``` ``` ``` (x - 32) dx + ``` ``` ``` ``` 489x + 2723 dx 85 + 18x + x2 ```

The left-hand piece is trivial to integrate:

 ``` ``` ``` (x - 32) dx = ``` ``` 1   2 ``` ``` x2 - 32x + C eq. 11.5-37 ```

But to do the other two pieces, you have to go to

Step 2: Complete the square. When you add half the middle coefficient to x you get the substitution of

 ``` v = x + 9 ``` and ```v2 = x2 + 18x + 81 ```

Clearly to get this back to the original denominator, you have to add 4 to v2:

```   v2 + 4  =  x2 + 18x + 85
```
and since  v = x + 9,  then  dv = dx.  But remember that you also have to put the  v = x + 9  substitution into the numerator of the remaining integral. To do this you have to convert the substitution to  v - 9 = x.  With these v substitutions the remaining integral becomes

 ``` ``` ``` 489x + 2723 dx = 85 + 18x + x2 ``` ``` ``` ``` 489(v - 9) + 2723 dv eq. 11.5-38a 4 + v2 ```

Step 3: Simplify the numerator. That is, multiply out the  489(v - 9)  term and then gather like terms. You have  489 × -9 = -4401  and  2723 - 4401 = -1678.  Using these results for the numerator and separating the sum into two integrals according to equation 11.2-13b, we get that equation 11.5-38a is the same as

 ``` ``` ``` 489v - 1678 dv = 4 + v2 ``` ``` 489 ``` ``` ``` ``` v dv - 1678 4 + v2 ``` ``` ``` ``` dv eq. 11.5-38b 4 + v2 ```

Step 4: Substitute. These are integrals you know how to do. The left-hand summand succumbs to simple substitution. For the one on the right, divide the 4 out of the denominator to get

 ``` -1678 ``` ``` ``` ``` dv = 4 + v2 ``` ``` 1678 -   4 ``` ``` ``` ``` dv eq. 11.5-39 1 + (v/2)2 ```

Then substitute

 ``` v tan(u) =   2 ``` and ``` (1 + tan2(u)) du = ``` ``` dv   2 ``` or ``` 2(1 + tan2(u)) du = dv ```

 ``` ``` ``` 3 - 2x - 14x2 + x3 = eq. 11.5-40 85 + 18x + x2 ```

 ``` 1   2 ``` ``` x2 - 32x + ``` ``` 489   2 ``` ``` ln|85 + 18x + x2| - ``` ``` 1678 arctan 2 ``` ``` ``` ```x + 9   2 ``` ``` ``` ``` + C ```

If you had to do the above example yourself from beginning to end on a homework problem or an exam, it would be naive to believe you wouldn't make any mistakes in arithmetic or carrying the correct signs through all the algebra. I, myself, will admit to having made a couple of mistakes the first time I ran through this one. But I caught them because at the end, I took the derivative of my answer, and it held manifest clues where I went wrong. I'll agree that taking the derivative of this answer and manipulating it back to the original integrand involves lots of fiddly work of putting things over common denominators and such. But it pays off in the end with a correct answer.

I have just one more trick to show you before we move on to hyperbolic substitution (which will involve again all the tricks you learned in this section). Once more we will find a way to algebraically munge something that we don't know how to integrate into something that we do. Suppose you wanted to integrate

 ``` ``` ``` 17 - 2x + 3x2 ______ dx eq. 11.5-41a √1 - x2 ```

You can't very well do polynomial long division on this quotient because the denominator is not a polynomial. The trick here is to make it a polynomial. And you do that by multiplying by

```    ______
√1 - x2
______  =  1
√1 - x2
```
As you know, multiplying by 1 doesn't change the function under the integral at all, but it does change its form:

 ``` ``` ``` 17 - 2x + 3x2 ______ dx = √1 - x2 ``` ``` ``` ``` 17 - 2x + 3x2   1 - x2 ``` ``` ______ √1 - x2 dx eq. 11.5-41b ```

```            -3
-x2 + 1 ) 3x2  -  2x  +  17
3x2  +  0x  -   3

-  2x  +  20
```
This gives a quotient of -3 and a remainder of  -2x + 20.  According to equation 11.5-32, our integral is the same as

 ``` ``` ``` 17 - 2x + 3x2   1 - x2 ``` ``` ______ √1 - x2 dx = ``` ``` ``` ``` ``` ``` -3 + ``` ``` -2x + 20   1 - x2 ``` ``` ``` ``` ______ √1 - x2 dx eq. 11.5-41c ```

You can now multiply the radical through using the distributive law and you have an integral of a sum. When you integrate a sum, you can apply equation 11.2-13b to make it into a sum of integrals:

 ``` ``` ``` ``` ``` -3 + ``` ``` -2x + 20   1 - x2 ``` ``` ``` ``` ______ √1 - x2 dx = ```
 ``` -3 ``` ``` ``` ``` ______ √1 - x2 dx - 2 ``` ``` ``` ``` x dx ______ + 20 √1 - x2 ``` ``` ``` ``` dx ______ eq. 11.5-41d √1 - x2 ```

We have done integrals of each of these three forms before. The middle one is easily done by simple substitution. The other two fall to a substitution of  sin(u) = x.  I will leave the details of the attack to you. At the end you should end up with

 ``` ``` ``` 17 - 2x + 3x2 ______ dx = √1 - x2 ```
 ``` 3 -   2 ``` ``` ______ x√1 - x2 + ``` ``` ______ 2√1 - x2 + ``` ``` 37 arcsin(x) + C eq. 11.5-41e 2 ```

(Pssst! Click here if you need a reminder on how to integrate the square of the cosine).

Note: We will be expanding the notion of trigonometric substitution in a later section. But before you get to that, we need to cover Partial Fractions first. In the mean time, all the integrands that you can integrate using the expanded trig substitutions will also yield to Hyperbolic Substitution, which is the next section. Some calculus courses do not cover hyperbolic functions. If you are in that boat, you can consider the treatment of them here to be optional and skip from here (after you do the exercise below) right to Partial Fractions. Or you might want to learn about hyperbolics anyway.

### Another Exercise

4) You're going to try a hard one now that will drill you on most of the stuff you have been learning in this unit as well as making you stretch to apply your knowledge to a form that is not quite the same as what you've seen before. Find the indefinite integral of

 ``` ``` ``` x2 ``` ``` ____________ √15 + 2x - x2 dx ```

1. Complete the square and make a substitution using the relationship that arises. I used the substitution variable, v, for this, so if you do the same, your development will look like mine. Note that you will have to substitute for that leading x2 term as well, which means turning your substitution around so that it expresses x in terms of your substitution variable.
2. It behooves you at this point to factor the constant out of the radical so that under the radical you have something in the form of  1 - (v/a)2,  where a is a constant. Pull that factor to outside the integral (remembering that when it passes from inside the radical to outside the radical, it gets square-rooted).
3. Look at the substituted integral that results and see how you can break it up into the sum of three separate integrals.
4. Examine each of the three separately and determine which method holds promise for integrating each piece. Note that for one of them it will not be trig substitution.
5. Determine the substitutions for each of the pieces. Carefully take the derivatives of each of the substitution equations to determine how you are going to substitute the dv in each piece. There is plenty of opportunity for sign errors or errors in multiplier-constants in this step. Take it slowly and keep your eyes open.
6. Make your substitutions and rewrite the three integrals in your new variables.
7. You will have to apply trig identities to whip the integrals you have at this point into a shape that you can integrate. One of them will require more than one trig identity application.
8. Integrate.
9. Before substituting back to v it will be necessary to apply trig identities again. And again one of the terms will require multiple applications.
10. Substitute the results back to an expression in v.
11. Last step -- substitute back to an expression in x.
Work through as much of this on your own as possible before looking at the solution. You might get all the way through. If you can't, go to the solution and follow along as far as you did get. From there do each step as recommended in the solution before clicking ahead to see the intermediate result. Then compare your intermediate result with mine.

If you work carefully and diligently, this problem will probably take you over an hour. If you then check it by taking the derivative of your answer, which is not trivial either, you will probably find that you made a mistake somewhere. Tracking down your mistake or mistakes will take more time. Be patient with yourself. Just for reference, this integral filled two full pages of my notebook to do and then check, not counting the pages I tore out and threw away on it.