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Answer Key for Exam 1
- Loosely speaking, this is the reciprocal of the second integral of the week, but this one is much easier: if we let u = x + sin x, then du = (1 + cos x)dx and we have ∫ 1 + cos x x + sin x dx =
du u = ln |u| + C = ln |x + sin x| + C.
2(a) This is #24 in the table with a = 1: ∫ dx x^2 + 1 = arctan x + C.
(b) If we substitute x = (^1) u in the above integral then dx = − duu 2 and we have
∫ dx x^2 + 1
− duu 2 1 u^2 + 1
− duu 2 1 u^2 + 1
u^2 u^2
du 1 + u^2
This is the same integral as in part (a) except for the minus sign, so we conclude that ∫ (^) dx x^2 + 1 = − arctan u + C = − arctan
x
+ C.
(c) Since the answers to (a) and (b) must be the same, it follows that
arctan x = − arctan^1 x
+ C
for some constant C. (Note: that’s what I was looking for in this part; what follows would be extra credit.) The constant isn’t exactly constant, however; there’s one constant when x > 0 and a different constant when x < 0. If we let x = 1 then we get
arctan 1 = − arctan 1 + C =⇒ π 4 = − π 4
and this is also what we’d get if we let x → ∞ or x → 0 +. But if we let x = −1 then we get
arctan(−1) = − arctan(−1) + C =⇒ − π 4
π 4
+ C =⇒ C = −
π 2
and this is also what we’d get if we let x → −∞ or x → 0 −. In conclusion,
arctan x + arctan
x =
{ (^) π 2 if^ x >^0 − π 2 if x < 0.
3(a) I hope you recognized this problem from the second lab. It becomes a lot easier if we first substitute x = w^2 to get rid of the square root. Then dx = 2w dw and we have
(1)
e
√x dx =
ew^2 w dw = 2
w ew^ dw.
Now this is a fairly easy integration by parts problem (alternatively, #14 in the table can be used with a = 1 and p(x) = x = w). We take u = 2w and dv = ew^ dw, so that du = 2 dw and v = ew^ and we have
(2) 2
w ew^ dw = 2
w ew^ −
ew^ dw
= 2 (w ew^ − ew) + C.
Combining this with (1) we have ∫ e
√x dx = 2 (w ew^ − ew) + C = 2
x e
√x − e
√x)
3(b) Substitute x = w^3 to get rid of the cube root. Then dx = 3w^2 dw and we have
(3)
e
√ (^3) x dx =
ew^3 w^2 dw = 3
w^2 ew^ dw.
Either use #14 in the table again, or integrate this by parts with u = w^2 and dv = ew^ dw, so that du = 2w dw and v = ew^ and we have
(4)
w^2 ew^ dw = w^2 ew^ − 2
w ew^ dw.
We did the remaining integral already in part (a), so putting (2), (3) and (4) together we have ∫ e
√ (^3) x dx = 3
[
w^2 ew^ − 2 (w ew^ − ew)
]
+ C
= 3ew^
w^2 − 2 w + 2
+ C
= 3e 3
√x ( x (^23) − 2 x (^13)
+ C.
- These integrals are all improper because the denominators become zero at the upper endpoint x = 1. Two of them can be done by elementary methods, while the other four need the gamma function (which was mentioned in problem 38 in section 7.7). The simplest one is (vi), because that’s the only one that the natural substitution u = 1 − x^6 works nicely on. We have du = − 6 x^5 dx, so that − du 6 = x^5 dx and
∫ √x^5 dx 1 − x^6
√du u
u−^ (^12) du = −
2 u
+ C = −
1 − x^6 + C.
Therefore (^) ∫ (^1)
0
x^5 dx √ 1 − x^6
1 − x^6
1
0
The other nice one is (iii). If we substitute u = x^3 then du = 3x^2 dx, so that du 3 = x^2 dx and we have
∫ x^2 dx √ 1 − x^6
du √ 1 − u^2
This we can look up; it’s #28 with a = 1. Therefore ∫ √x^2 dx 1 − x^6
=^1
arcsin u + C =^1 3 arcsin
x^3
+ C,
and so (^) ∫ 1 0
x^2 dx √ 1 − x^6
arcsin
x^3
1
0
(arcsin 1 − arcsin 0) =
( (^) π 2
π 6
For the record, if we define
G =
[
3
)] 3
2 43 π with Γ(x) denoting the gamma function, then
If we fully multiply out the right side of (6) we get
x = (A + C)x^3 + (A + B + 6C + D)x^2 + (9A + 9C + 6D)x + 27A + 9B + 9D,
and equating coefficients of the various powers of x gives
(i) A + C = 0, (ii) A + B + 6C + D = 0, (iii) 9A + 9C + 6D = 1, (iv) 27A + 9B + 9D = 0.
Assuming we know already that B = − 16 but nothing else, we can get D by multiplying (i) by 9 and substituting in (iii), which gives 6D = 1, D = 16. Therefore B + D = 0, and using this in (iv) we get 27 A = 0, so that A = 0. Then (i) or (ii) implies that C = 0.
Once we know A, B, C, D we have
∫ (^) ∞
0
x dx (x + 3)^2 (x^2 + 9)
0
x + 3
(x + 3)^2
0 x + (^16) x^2 + 9
dx =
0
dx x^2 + 9
0
dx (x + 3)^2
which was (5), so with more or less difficulty we are eventually led to (5). We can look up the first integral in (5); using #24 with a = 3 we have
1 6
0
dx x^2 + 9
arctan x 3
∞
0
(arctan ∞ − arctan 0) =
( (^) π 2
π 36
The second integral in (5) we can do straightaway:
1 6
0
dx (x + 3)^2
x + 3
∞
0
Or, if you’re not sure about this integration, you can substitute u = x + 3, so that du = dx. If x = 0 then u = 0 + 3 = 3, and if x = ∞ then u = ∞ + 3 = ∞, so this gives
1 6
0
dx (x + 3)^2
3
du u^2
3
u−^2 du
u−^1 − 1
∞
3
6 u
∞ 3 =
18 =^
So we finally have
∫ (^) ∞
0
x dx (x + 3)^2 (x^2 + 9)
=^1
0
dx x^2 + 9
0
dx (x + 3)^2 = π 36
= π^ −^2 36
