Math 222 - Worksheet 3 Solutions
Name:
Z
1. Find
√
x 2x + 5dx.
Solution: The important thing here is that both the terms outside the square root and inside
the square root are linear. So, we let u = 2x + 5, and we can easily rewrite x = u−5
2 . So,
dx = 21 du. We get:
Z
2. Find
Z p
√
x 2x + 5dx =
u − 5√ 1
u du
2
2
Z
√
1
=
(u − 5) udu
4
1 2 5/2 10 3/2
=
u − u
+c
4 5
3
1
10
= (2x + 5)5/2 − (2x + 5)3/2 + c
10
3
Z
x2 + 6x + 8dx.
Solution: We first complete the square. Note that x2 + 6x + 8 = (x + 3)2 − 1. So, the integral
becomes:
Z p
(x + 3)2 − 1dx
We could do a trigonometric substitution here, but that reduces to finding the integral of
tan(θ) sec2 (θ)dθ. This is not simple (though it can be done, using the integrals of sec(θ) and
sec3 (θ)). Instead, we use rational substitution.
Recall U (t) = 21 (t + 1t ), V (t) = 12 (t − 1t ). Then U 2 = V 2 + 1. So, we do the substitution x + 3 = U ,
so x = U − 3, and so dx = U 0 dt. So, the integral becomes:
Z p
Z
Z
1
1 1
1
0
0
2
U − 1U dt = V U dt =
(t − ) (1 − 2 )dt
2
t 2
t
Z
1
2
1
=
t − + 3 dt
4
t
t
1 2 11
1
= t − 2 − ln |t| + c
8
8t
4
1
2
Remember that t = U + V, 1t = U − V . So, substituting this in, we get our integral equals:
1
1
1
1
1
(U + V )2 − (U − V )2 − ln |U + V | + c = (U 2 + 2U V + V 2 − (U 2 − 2U V + V 2 )) − ln |U + V | + c
8
8
4
8
4
1
1
= U V − ln |U + V | + c
4
4
Finally, U = x + 3, and V =
√
U2 − 1 =
p
(x + 3)2 − 1, so our final answer is:
p
p
1
1
(x + 3) (x + 3)2 − 1 − ln |x + 3 + (x + 3)2 − 1| + c
4
4
Z
1
3. Consider
2x2 dx. Use the substitution u = x2 , du = 2xdx, and change the limits of integration
−1
to be in terms of u. What answer do you get? Is this correct? Explain.
Solution: Letting u = x2 , we know that at x = −1, u = 1 and at x = 1, u = 1. So, the bounds
of the integral
R a change to 1 and 1, so the integral is 0. This is by the Fundamental Theorem of
Calculus, a f (x)dx = 0 for any integrable function f (x).
This is not correct. By direct computation, we have:
Z
1
2 3
2x dx =
x
3
−1
2
1
=
−1
2 2
4
+ =
3 3
3
The reason our substitution got the wrong answer has to do with the domain x is defined on. We
√
are integrating over [−1, 1]. But, if x = u, then x must be ≥ 0. So, this substitution doesn’t
work on the domain [−1, 1]. It does work on the domain [0, 1], however, and in fact the following
is true:
Z 1
Z 1
2
2x dx = 2
2x2 dx
−1
0
The substitution works for the integral on the right, so we could compute the integral that way.
Z
4. Find sec(x)dx in two different ways:
(a) Multiply by
sec(x) + tan(x)
and use u-substitution.
sec(x) + tan(x)
(b) Note that sec(x) =
cos(x)
. Manipulate this further, and then use u-substitution and
cos2 (x)
partial fractions.
Solution 1: We do as the hint says to get:
Z
Z
sec(x) + tan(x)
dx
sec(x)dx = sec(x)
sec(x) + tan(x)
Z
sec2 (x) + sec(x) tan(x)
=
dx
sec(x) + tan(x)
3
Notice that the derivative of the denominator is the numerator. Letting u = sec(x) + tan(x) on
the domain − π2 < x < π2 , we get:
Z
Z
du
sec2 (x) + sec(x) tan(x)
dx =
sec(x) + tan(x)
u
= ln |u| + c = ln | sec(x) + tan(x)| + c
Solution 2: We do as the second hint says to get:
Z
Z
Z
cos(x)
cos(x)
sec(x)dx =
dx =
dx
cos2 (x)
1 − sin2 (x)
We do a substitution u = sin(x), so du = cos(x) getting:
Z
Z
cos(x)
du
dx
=
2
1
−
u2
1 − sin (x)
Note that the denominator now factors as (1 + u)(1 − u). We use partial fractions to rewrite as
A
B
1+u + 1−u . By the heavyside method, we have:
• A=
1
=
1 − (−1)
1
2
1
= 12
1+1
Our integral is now:
Z
• B=
1
1
1
1
+
du = ln |1 + u| − ln |1 − u| + c
2(1 + u) 2(1 − u)
2
2
1
1
= ln |1 + sin(x)| − ln |1 − sin(x)| + c
2
2
NOTE: It seems we have two different answers now. In fact, they are the same. Using properties
A
of logarithms (specifically, ln( B
) = ln(A) − ln(B) and y ln(A) = ln(Ay )) we have the following:
1
1 1 + sin(x) 1
ln |1 + sin(x)| − ln |1 − sin(x)| = ln 2
2
2
1 − sin(x) 1 1 + sin(x) 1 + sin(x) = ln 2
1 − sin(x) 1 + sin(x) 1 (1 + sin(x))2 = ln 2
1 − sin2 (x) 1 (1 + sin(x))2 = ln 2
cos2 (x) 1 + sin(x) = ln cos(x) = ln | sec(x) + tan(x)|
Z
5. (a) Find
sec3 (x)dx. (Hint: There are many ways to do this. One approach is to use integra-
tion by parts and the exercise above.)
4
Z
(b) Given that you can integrate
3
sec (x)dx, how would this help you find
Z p
x2 + 36dx?
Solution: Note that sec3 (x) = sec2 (x) sec(x). Since we can easily integrate sec2 (x), and we
can easily differentiate sec(x), we let F = sec(x), G0 = sec2 (x) in integration by parts. Then
F 0 = sec(x) tan(x), G = tan(x), so we get:
Z
Z
sec3 (x)dx = sec(x) tan(x) − sec(x) tan2 (x)dx
Z
= sec(x) tan(x) − sec(x)(sec2 (x) − 1)dx
Z
Z
= sec(x) tan(x) − sec3 (x)dx + sec(x)dx
Moving the
R
sec3 (x)dx to the other side, we get:
Z
Z
2 sec3 (x)dx = sec(x) tan(x) + sec(x)dx
= sec(x) tan(x) + ln | sec(x) + tan(x)| + c
This last step was by the previous exercise. Dividing by 2, our final answer is:
Z
1
1
sec3 (x)dx = sec(x) tan(x) + ln | sec(x) + tan(x)| + c
2
2
Z
6. Find
√
x2
dx
.
+ 2x + 2
Solution:
We complete the square. So, x2 + 2x + 2 = (x + 1)2 + 1. Since we are now concerned
p
with (x + 1)2 + 1, it makes sense to do the substitution x + 1 = tan(θ), − π2 < θ < π2 . So,
dx = sec2 (θ)dθ.
We get:
Z
Z
dx
p
=
(x + 1)2 + 1
sec2 (θ)dθ
dθ
sec(θ)
Z
=
sec(θ)dθ
= ln | sec(θ) + tan(θ)| + c
We could substitute in x + 1 = tan(θ) → θ = arctan(x + 1) now and be done, or we could
do the following.
We know
p
p x + 1 = tan(θ), so we just need to determine what sec(θ) is. But,
2
sec(θ) = tan (θ) + 1 = (x + 1)2 + 1. Therefore, the final answer is:
Z
p
dx
√
= ln | (x + 1)2 + 1 + x + 1| + c
x2 + 2x + 2
5
Z
7. Use two different methods to find
x2
x+4
dx.
+ 8x + 12
Solution 1: Since this is a rational function, we can certainly use partial fractions. Note that
x2 + 8x + 12 = (x + 2)(x + 6). So, we should have:
x2
x+4
A
B
=
+
+ 8x + 12
x+2 x+6
By the heavyside method, we have:
2
x + 4 1
= =
• A=
x + 6 x=−2 4
2
−2
x + 4 1
=
• B=
=
x + 2 x=−6 −4
2
So, we have:
Z
x+4
=
x2 + 8x + 12
Z
1
1
+
dx
2(x + 2) 2(x + 6)
1
1
= ln |x + 2| + ln |x + 6| + c
2
2
d 2
(x + 8x + 12) = 2x + 8 = 2(x + 4).
dx
2
So, doing the substitution u = x + 8x + 12, du = 2(x + 4)dx so we get:
Z
Z
1 du
x+4
dx =
2
x + 8x + 12
2 u
1
= ln |u| + c
2
1
= ln |x2 + 8x + 12| + c
2
Solution 2: A somewhat simpler solution is to notice that
Z
8. Find
1
√
dx.
1+ x+1
Solution: This integral involves a non-obvious substitution. The idea here is that the
√ square
root in the denominator makes things difficult. So, we do the substitution u = 1 + x + 1 to
make things look nicer, and see where it takes us. This is common in integration. Oftentimes
you just have to try things and see where they lead.
√
Since u = 1 + x + 1, x = (u − 1)2 − 1 = u2 − 2u, so dx = (2u − 2)du. We then get:
Z
2u − 2
du
u
Z
u−1
=2
du
u
Z
1
= 2 (1 − )du
u
= 2 u − ln |u| + c
√
√
= 2(1 + x + 1) − 2 ln |1 + x + 1| + c
dx
√
=
1+ x+1
Z