Karl's Calculus Tutor - Solutions to Problems in 11.3

# Solutions to Problems in 11.3

### Solution to Exercise 1:

The problem was to use simple substitution to find the indefinite integral of

 ``` ``` ``` x e-x2/2 dx ```
Step 1) What part of the integrand is the derivative of another part, and is that derivative a factor of the integrand? The x up front is the derivative of the  x2/2  that appears inside the exponential. That up-front x also meets the second requirement -- that it be a factor of the integrand (that is it is multiplied by the rest of the integrand). So at this point you've identified the two pieces of the integrand you need in order to proceed.

Step 2) Prepare your substitution variable. In step 1, you identified the up-front x as being the derivative of the  x2/2.  So let  u = x2/2.  Then

```   du
=  x
dx
```
and when you multiply through by dx you find that  du = x dx

Step 3) Make the substitution. Remember the tip about multiplication being commutative. That means that this integral is the same as

 ``` ``` ``` e-x2/2 x dx ```
So now  x dx  becomes du and  x2/2  becomes u. Your substituted integral is
 ``` ``` ``` e-u du ```

Step 4) Integrate the substituted integral. Use the rules from the table and equation 11.2-9 to help you with this. You ought to get

 ``` ``` ``` e-u du = -e-u + C ```

Step 5) Substitute back. Your original substitution was  u = x2/2.  So put that back into the expression you got by integrating the substituted integral.

 ``` ``` ``` e-u du = -e-u + C = -e-x2/2 + C ```
That is your answer. Now take the derivative of your answer (using the chain rule) to verify that you get back the original integrand.

# Solutions to Problems in 11.3

### Solution to Exercise 2

The problem was to use simple substitution of variables to find the indefinite integral of

 ``` ``` ``` __________ sin(θ) √1 - cos(θ) dθ ```
Step 1) What part of the integrand is the derivative of another part? Clearly  sin(θ)  is the derivative of  -cos(θ).  But if you substituted  u = -cos(θ),  you'd have some stuff left over under the radical after you made the substitution. Observe that  sin(θ)  is also the derivative of  1 - cos(θ),  and by substituting  u = 1 - cos(θ)  you will end up with more simplification.

Step 2) Prepare your substitution variable. If we make the substitution,  u = 1 - cos(θ),  then

```  du
=  sin(θ)
dθ
```
and when you multiply through by dθ, you get
```   du  =  sin(θ) dθ
```

Step 3) Make the substitution. Remember about the commutativity of multiplication that enables you to move the  cos(θ)  so that it is next to the dθ. When you do that and make the substitution, you will get

 ``` ``` ``` _ √u du ```

 ``` ``` ``` _ √u du = ``` ``` 2   3 ``` ``` u3/2 + C ```

Step 5) Substitute back. Your original substitution for this was  u = 1 - cos(θ).  So substitute back for u

 ``` ``` ``` _ 2 √u du =   3 ``` ``` u3/2 + C = ``` ``` 2   3 ``` ``` (1 - cos(θ))3/2 + C ```
which is the answer. Finally, you get to do the step that automatically follows -- that is verifying your answer by matching its derivative with the original integrand.

# Solutions to Problems in 11.3

### Solution to Exersice 3

The problem was to find the indefinite integral of

 ``` ``` ``` cos3(θ) dθ ```
The hint told you to use the identity,  cos2(θ) = 1 - sin2(θ).  To do that you have to find a  cos2(θ)  in the integral itself. So the first thing to do here is

Step 0) Convert the integrand to a form on which substitution will work.

 ``` ``` ``` cos3(θ) dθ = ``` ``` ``` ``` cos(θ) cos2(θ) dθ = ``` ``` ``` ``` cos(θ) (1 - sin2(θ)) dθ ```
The hint also recommended that you should break this into the difference of two separate integrals.
 ``` ``` ``` cos(θ) (1 - sin2(θ)) dθ = ``` ``` ``` ``` cos(θ) dθ - ``` ``` ``` ``` cos(θ) sin2(θ) dθ ```
As predicted in the hint, the first integral in the difference is easy. For the second you need to

Step 1) Find what part of the integrand is the derivative of another part. Clearly  cos(θ)  is the derivative of  sin(θ).  So that will form the basis of the substitution for the second integral in the difference.

Step 2) Prepare your substitution variable. If you substitute  u = sin(θ),  then

```   du
=  cos(θ)
dθ
```
Then multiplying through by dθ, you get
```  du  =  cos(θ) dθ
```
Now, in the second integrand, move the  cos(θ)  next to the dθ (which you can do by commutativity of multiplication), and you're ready to

Step 3) Make the substitution. Remember that you are substituting only the right-hand integral of the difference. Leave the left-hand integral in the difference alone. In the right-hand integral, substitute u for  sin(θ)  and du for  cos(θ) dθ

 ``` ``` ``` cos(θ) dθ - ``` ``` ``` ``` sin2(θ) cos(θ) dθ = ``` ``` ``` ``` cos(θ) dθ - ``` ``` ``` ``` u2 du ```

Step 4) Integrate. The left-hand integral in the difference is easy and you can find it on the table. The right-hand integral succumbs to equation 11.2-4f.

 ``` ``` ``` cos(θ) dθ - ``` ``` ``` ``` u2 du = sin(θ) - ``` ``` 1   3 ``` ``` u3 + C ```
(Again we combined the two undetermined constants from the two integrals into a single C)

Step 5) Substitute back. Your original substitution was  u = sin(θ).  So put  sin(θ)  in everywhere you see a u. You get

 ``` sin(θ) - ``` ``` 1   3 ``` ``` sin3(θ) + C ```
which is the answer. Take its derivative and see how you can manipulate that back into the original integrand.

# Solutions to Problems in 11.3

### Solution to Exercise 4

The problem was to use simple substitution to integrate

 ``` ``` ``` 2θ cos(θ2) sin(θ2) dθ ```
Step 1) What part of the integrand is the derivative of another part? If you had remarkably keen perception, you might have observed that  2θ cos(θ2)  is the derivative of  sin(θ2).  But those who have progressed far enough to see that probably don't need to be reading this tutorial. Most folks, though, would see that  2θ  is the derivative of  θ2.  So we'll go with that and see where it leads.

Step 2) Prepare your substitution variable. If you are going to substitute  u = θ2,  then

```   du
=  2θ
dθ
```
And when you multiply through by dθ, you get
```   du  =  2θ dθ
```
When you move the  2θ  in the original integrand so that it is proximate to the dθ, you find that you are ready to

Step 3) Make the substitution. So you substitute θ2 with u and  2θ dθ  with du.

 ``` ``` ``` cos(θ2) sin(θ2) (2θ) dθ = ``` ``` ``` ``` cos(u) sin(u) du ```
The new integrand is simpler than the one you started with but still presents difficulty. Remember the hint said that sometimes once is not enough. This new integrand is susceptible to further substitution. Why? Because  cos(u)  is the derivative of  sin(u).  So if you let  v = sin,  then
```  dv
=  cos(u)
du
```
and multiplying through by du you get
```   dv  =  cos(u) du
```
Now substitute again -- this time replace  sin(u)  with v and  cos(u) du  with dv.
 ``` ``` ``` cos(θ2) sin(θ2) (2θ) dθ = ``` ``` ``` ``` cos(u) sin(u) du = ``` ``` ``` ``` v dv ```
Step 4) Integrate the substituted integrand. This one is pretty easy. Remember that  v = v1,  and apply equation
11.2-4f. You find that
 ``` ``` ``` 1 v dv =   2 ``` ``` v2 + C ```
Step 5) Substitute back. Since you substituted twice, you will have to substitute back twice -- in the reverse order that you substituted going in. The last substitution you made was  v = sin(u).  So that is the first back substitution to make:
 ``` ``` ``` 1 v dv =   2 ``` ``` v2 + C = ``` ``` 1   2 ``` ``` sin2(u) + C ```
Now back substitute using the first of your substitutions, which was  u = θ2
 ``` ``` ``` 1 v dv =   2 ``` ``` v2 + C = ``` ``` 1   2 ``` ``` sin2(u) + C = ``` ``` 1   2 ``` ``` sin2(θ2) + C ```
To verify this answer by taking the derivative, you will have to employ the chain rule twice.

PS) If you did have the keen perception to see that  2θ cos(θ2)  is the derivative of  sin(θ2),  you would have been able to do this integral problem with just a single substitution instead of having to substitute twice. Now that you know, go ahead and try doing that way.

# Solutions to Problems in 11.3

### Solution to Exercise 5

The problem was to use substitution of variables to integrate

 ``` ``` ``` (3x2 - 6x + 2) dx   x3 - 3x2 + 2x - 9 ```
Step 1) What part of the integrand is the derivative of another part? In this one the numerator expression (sans the dx) is the derivative of the denominator expression. We'll work this one through the remaining steps, and then I'll have a comment about the general case of when the numerator is the derivative of the denominator.

Step 2) Prepare your substitution variable. You will be substituting the entire denominator. So you will have  u = x3 - 3x2 + 2x - 9.  Then

```  du
=  3x2 - 6x + 2
dx
```
and multiplying through by dx, you get
```  du  =  (3x2 - 6x + 2) dx
```

Step 3) Make the substitution. So you will substitute  u = x3 - 3x2 + 2x - 9  with u and  (3x2 - 6x + 2) dx  with du.

 ``` ``` ``` (3x2 - 6x + 2) dx = x3 - 3x2 + 2x - 9 ``` ``` ``` ``` du   u ```
Step 4) Integrate the substituted integral. If you don't remember how to integrate this, see equation
11.2-6c.
 ``` ``` ``` (3x2 - 6x + 2) dx = x3 - 3x2 + 2x - 9 ``` ``` ``` ``` du = ln|u| + C u ```
Step 5) Substitute back. Your original substitution was  u = x3 - 3x2 + 2x - 9,  which is what you use to substitute back.
 ``` ``` ``` (3x2 - 6x + 2) dx = x3 - 3x2 + 2x - 9 ``` ``` ``` ``` du = ln|u| + C = ln|x3 - 3x2 + 2x - 9| + C u ```
The expression on the right is your answer. Take its derivative using the chain rule to verify it.

Something to remember: Whenever you have a quotient that you need to integrate, and the numerator is exactly equal to the derivative of the denominator, the integral will always be equal to the natural log of the denominator. That is, if you have

 ``` ``` ``` f(x) dx   g(x) ```
and
```           dg
f(x)  =
dx
```
then
 ``` ``` ``` f(x) dx = ln|g(x)| + C g(x) ```
If you take the derivative of  ln|g(x)|,  you will see why. Another way to look at it is this:
 ``` ``` ``` f(x) dx = g(x) ``` ``` ``` ``` dg dx dx = g ``` ``` ``` ``` dg = ln|g| + C g ```
Remember in the main text when we found the integral of  cot(θ) dθ?  That was an example of exactly this phenomenon. Can you use a slight variation on it to find
 ``` ``` ``` tan(θ) dθ ```

# Solutions to Problems in 11.3

### Solution to Exercise 6

The problem was to use substitution of variables to integrate

 ``` ``` ``` ______ x √3x - 7 dx ```
Step 1) Observe that the stuff under the radical is a linear function of x. That means that if you use that as your substitution function (i.e.  u = 3x - 7),  you will be able to solve for x in terms of u easily.

Step 2) Prepare your substitution variable. If you substitute  u = 3x - 7,  then

```   du
=  3
dx
```
and
```   du  =  3 dx
```
To get a  3 dx  in the integrand, you will have to multiply it by 3/3 and pull the denominator outside of the integral:
 ``` 1   3 ``` ``` ``` ``` ______ x √3x - 7 (3) dx ```
And in order to substitute that x to the left of the radical you also need to solve for x in terms of u.
```  u  =  3x - 7
```
so
```        u + 7
x  =
3
```

Step 3) Make the substitution. You are substituting u for  3x - 7,  and du for  3 du.  And where the x appears by itself, you are going to substitute it with  (u + 7)/3

 ``` 1   3 ``` ``` ``` ``` ______ x √3x - 7 (3) dx = ``` ``` 1   3 ``` ``` ``` ```u + 7   3 ``` ``` ``` ``` _ √u du = ``` ``` 1   9 ``` ``` ``` ``` (u + 7) u1/2 du ```

Step 4) Multiply through and then break it into two the sum of two integrals. That gives you

 ``` 1   9 ``` ``` ``` ``` (u + 7) u1/2 du = ``` ``` 1   9 ``` ``` ``` ``` (u3/2 + 7u1/2) du = ``` ``` 1   9 ``` ``` ``` ``` u3/2 du + ``` ``` 7   9 ``` ``` ``` ``` u1/2 du ```

Step 5) Integrate. Use equation 11.2-4f to integrate these two summands:

 ``` 1   9 ``` ``` ``` ``` u3/2 du + ``` ``` 7   9 ``` ``` ``` ``` u1/2 du = ``` ``` 2   45 ``` ``` u5/2 + ``` ``` 14   27 ``` ``` u3/2 + C ```

Step 6) Substitute back. Your original substitution function was  u = 3x - 7.  So

 ``` 2   45 ``` ``` u5/2 + ``` ``` 14   27 ``` ``` u3/2 + C = ``` ``` 2   45 ``` ``` (3x - 7)5/2 + ``` ``` 14   27 ``` ``` (3x - 7)3/2 + C ```
Now take the derivative of this answer and do the appropriate algebraic munging to verify that you can get back to the original integrand.

# Solutions to Problems in 11.3

### Solution to Exercise 7

The problem was to use simple substitution of variables to find the indefinite integral of

 ``` ``` ``` dx   2(√x + 1 + 7) √x + 1 ```

Step 1) What part of the integrand is the derivative of another part? The hint mentioned that when you take the derivative of a square root you get one half the reciprocal of the square root. The observation to make on this one is that

```     1

2√x + 1
```
is the derivative of
```   _____
√x + 1 + 7
```

Step 2) Prepare your variables for substitution. If you let

```          _____
u  =  √x + 1 + 7
```
then
```   du        1
=
dx     2√x + 1
```
and multiplying through by dx you get
```             dx
du  =
2√x + 1
```

Step 3) Make the substitution. You have to use the commutative law of addition to move the 2 in the denominator so that it groups with the last factor in the denominator, and when you do, the substitution is perfect:

 ``` ``` ``` dx   = 2(√x + 1 + 7) √x + 1 ``` ``` ``` ``` du   u ```

 ``` ``` ``` du = ln|u| + C u ```

Step 5) Substitute back. Your original substitution was

```         _____
u  =  √x + 1 + 7
```
So put that expression in for u and you get
 ``` ``` ``` dx   = 2(√x + 1 + 7) √x + 1 ``` ``` ``` ``` du = ln|u| + C = u ``` ``` _____ ln|√x + 1 + 7| + C ```
which is the answer. As always, take the derivative of this using the chain rule to verify that you get back the original integrand. When you're done with that, ask yourself, if this same integral problem had been presented in the equivalent form of
 ``` ``` ``` dx   2x + 2 + 14√x + 1 ```
would you have figured out that you had to factor the
```   _____
2√x + 1
```
out of the denominator to get it into a form where simple substitution would work? Look carefully at how that works, and remember that often you will have to do algebraic manipulations to put an integrand into a form that you know how to integrate.