Download 5.6: 18 5.6: 32 5.6: 52

5.6: 18
Use the substitution formula in Theorem 7 to evaluate the integral:
3π/2
Z
cot5 (θ/6) sec2 (θ/6) dθ.
π
Sometimes its just best to simplify first. Recall that cot x = cos x/ sin x, and that sec x = 1/ cos x. Then we
can reduce this to:
Z 3π/2
Z 3π/2
−1
5
2
3
2
cot (θ/6) sec (θ/6) dθ = −6
cot (θ/6)
csc (θ/6) dθ
6
π
π
d
cot x = − csc2 x. Then we see that we should let u = cot(θ/6), whereby du = − 16 csc2 (θ/6) dθ.
Recall that dx
Then upon the use of the substitution formula:
1
Z 3π/2
Z u(3π/2)
−1
u4 3
2
3
−6
cot (θ/6)
= 12.
csc (θ/6) dθ = −6
u du =
6
4 √3
π
u(π/2)
5.6: 32
Use the substitution formula in Theorem 7 to evaluate the integral:
Z 16
dx
√
.
2x ln x
2
Here we let u = ln x, so du =
1
x
Z
dx:
16
2
dx
1
√
=
2
2x ln x
Z
u
ln 16
(2)u(16) u−1/2 du = u1/2 .
ln 2
4
Before simplifying further, notice that ln 16 = ln 2 = 4 ln 2. Then:
ln 16
√
√
√
1/2 = 4 ln 2 − ln 2 = ln 2.
u ln 2
5.6: 52
Find the area shown in the book.
The formula for the area will be:
π/3
Z
A=
−π/3
1
2
2
sec t − −4 sin t dt
2
Simplifying a little bit, we get:
A=
1
2
Z
π/3
sec2 t dt + 4
Z
π/3
−π/3
−π/3
1
sin2 t dt.
The first integral can be evaluated as follows:
Z
π/3
π/3
√
sec t dt = tan t
= 2 3.
2
−π/3
−π/3
The second integral requires the use of a half angle formula:
Z
π/3
sin2 t dt =
−π/3
1
2
Z
π/3
(1 − cos 2t) dt =
−π/3
√ 1 4π − 3 3 .
12
Combining these results, we get:
A=
√ 4π
1 √ 1 2 3 +4
4π − 3 3 =
.
2
12
3
5.6: 100
Find the area of the “triangular” region in the first quadrant that is bounded above by the curve y = ex/2 ,
below by the curve y = e−x/2 , and on the right by the line 2 ln 2.
Be sure that, for a problem of this form, you can sketch the given regions and understand what they look
like. The area is given by:
Z
2 ln 2
A=
e
x/2
−e
−x/2
dx = 2e
x/2
+ 2e
−x/2
0
2 ln 2
1
= 2 2 − 1 + − 1 = 1.
2
0
5.6: 114
(a) Show that if f is odd on [−a, a], then
Z
a
f (x) dx = 0.
−a
(b) Test the result in part (a) with f (x) = sin x and a = π/2.
I’ll verify part (a). Part (b) should be fairly straightforward. Let’s try to compute
substitution u = −x, du = −dx:
Z
0
Z
u(0)
f (x) dx =
−a
Z
u(−a)
0
a
−a
f (x) dx by using the
a
Z
f (u) du = −
f (−u) (−du) =
R0
f (u) du.
0
The second to last step follows from the fact that since f is an odd function, f (−u) = −f (u), and the last
step from inverting the order of the bounds. Now u is just a dummy index, so we can relabel it x if we like.
Then:
Z a
Z 0
Z a
Z a
Z a
f (x) dx =
f (x) dx +
f (x) dx = −
f (x) dx +
f (x) dx = 0.
−a
−a
0
0
2
0