Math 105/206 - Practice problems for Quiz 3
Problem 1
Compute the following trigonometric indefinite integrals:
Z
Z
3
5
sin x cos x dx
sin2 x cos2 x dx
Z
sec−2 x tan3 x dx
Solution
Split off a sine from the product sin3 x cos5 x to get (sin x)(sin2 x cos5 x) and use sin2 x = 1 − cos2 x:
Z
Z
3
5
sin x cos x dx = sin x(1 − cos2 x) cos5 x dx.
Now use the substitution u = cos x, so du = − sin x dx, and we get
Z
Z
2
5
sin x(1 − cos x) cos x dx = −(1 − u2 )u5 du.
Now you can easily integrate this and plug in u = cos x in the result to get
Z
cos6 x cos8 x
+
+C
sin3 x cos5 x dx = −
6
8
For the second integral use sin2 x = 1 − cos2 x to get
Z
Z
Z
Z
2
2
2
4
2
sin x cos x dx = (cos x − cos x) dx = cos x dx − cos4 x dx
and now compute the two integrals by applying the half-angle formula
1 + cos(2x)
2
(you will have to apply it twice for the second integral, the second time to write cos2 2x in terms of cos(4x), we
saw this in class).
Your final result should be
Z
x sin(4x)
+ C.
sin2 x cos2 x dx = −
8
32
2
Alternatively you can write sin2 x cos2 x = sin(2x)
and go on from here.
2
cos2 x =
sin x
For the third integral, remember that tan x = cos
and sec x = cos1 x to get
x
Z
Z
sin3 x
−2
3
sec x tan x dx =
dx.
cos x
Now split off a sine and use sin2 x = 1 − cos2 x
Z
Z
sin3 x
1 − cos2 x
dx = sin x ·
dx
cos x
cos x
and use substitution with u = cos x to get
Z
Z
1 − cos2 x
1 − u2
sin x ·
dx = −
du.
cos x
u
You can compute this as usual. The final result is
Z
cos2 x
sec−2 x tan3 x dx = − log | cos x| +
+C
2
Problem 2
Use trigonometric substitutions to compute the following indefinite integrals:
Z √
Z
dx
2
9 − 4x dx
2
x − 6x + 34
Solution
2
2
For
q the first integral,
q we almost have an expression like a −x in the integrand. To get it, just write
4( 94 − x2 ) = 2 94 − x2 , so a = 23 .
√
9 − 4x2 =
The recipe tells us to use x = 23 sin u (so u = arcsin 23 x and u is between − π2 and π2 ), so dx = 32 cos u du. We
get
Z
r
2
9
− x2 dx = 2
4
Z r
3
9
9
cos2 u · cos u du = 2 · ·
4
2
4
Z
| cos u| · cos u du
Now | cos u| = cos u since u ∈ [− π2 , π2 ], and integrating cos2 u as usual you get to
Z √
9
9
9 − 4x2 = u + sin(2u).
4
8
Now u = arcsin 32 x, and sin(2u) = 2 sin u cos u, and drawing a right triangle with one angle that is u, you see
q
that cos u = 1 − 49 x2 , so
r
Z √
9
2
9
2
4
9 − 4x2 = arcsin x + · x · 1 − x2
4
3
4 3
9
For the second integral, complete the square in the denominator and write it as (x − 3)2 + 25, perform the
substitution u = x − 3 to get
Z
Z
dx
du
=
.
2
2
x − 6x + 34
u + 52
This integral is one of the basic ones
Z
1
u
du
= arctan + C
2
2
u +a
a
a
so
Z
x2
1
x−3
dx
= arctan
+C
− 6x + 34
5
5
Problem 3
Use the method of partial fractions to compute the following indefinite integral
Z
dx
3
2
x − 2x − 4x + 8
Solution
The denominator factors as
x3 − 2x2 − 4x + 8 = x2 (x − 2) − 4(x − 2) = (x2 − 4)(x − 2) = (x − 2)2 (x + 2)
so we are looking for a partial fractions decomposition
x3
−
1
A
B
C
=
+
+
.
2
− 4x + 8
x − 2 (x − 2)
x+2
2x2
By using a common denominator in the right-hand side we get
A
B
C
A(x − 2)(x + 2) + B(x + 2) + C(x − 2)2
+
+
=
x − 2 (x − 2)2 x + 2
(x − 2)2 (x + 2)
so we have to solve
1 = A(x − 2)(x + 2) + B(x + 2) + C(x − 2)2
for A, B, C.
1
1
By expanding and solving the equations we get A = − 16
, B = 14 , C = 16
.
By using these and integrating
Z 1
1
1
−
+
dx
+
16(x − 2) 4(x − 2)2 16(x + 2)
we find
Z
x3
−
dx
1
1
1
1
= − log |16(x − 2)| − ·
+
log |16(x + 2)|
− 4x + 8
16
4 x − 2 16
2x2
Problem 4
Compute the following improper integrals
Z ∞
2−x dx
Z
1
3
4
dz
(z − 3)3/2
Solution
Let us compute
Z
2−x dx.
Since 2−x = e−x·loge 2 , we get
Z
−x
2
For any a > 1 we compute
Z a
2−x dx =
1
Z
dx =
e−x loge 2 dx =
a
1
1
· e−x loge 2 =
(e−a loge 2 − e− loge 2 ).
− loge 2
−
log
2
e
1
Now as a → ∞, we have e−a loge 2 → 0, so
Z ∞
2−x dx =
1
1
· e−x loge 2 + C
− loge 2
1
1
(−e− loge 2 ) =
− loge 2
2 loge 2
For the second, let us compute
Z
dz
(z − 3)3/2
by using the substitution u = z − 3. We find
Z
Z
dz
du
2
2
=
= −√ + C = −√
+C
3/2
3/2
(z − 3)
u
u
z−3
Consequently, for any 3 < a < 4 we have
4
Z 4
dz
2 2
= −√
= −2 + √
3/2
z−3 a
a−3
a (z − 3)
and as a → 3, this diverges.
Problem 5
Find the trapezoid rule approximations for the integral
Z 9
x2 dx
1
using n = 2 and n = 4 subintervals.
Solution
Set f (x) = x2 . For n = 2, the two subintervals are [1, 5] and [5, 9], so ∆x = 4 and we get
f (1) + f (5) f (5) + f (9)
+
= 4(13 + 53) = 264
T (2) = 4
2
2
For n = 4, the four subintervals are [1, 3], [3, 5], [5, 7] and [7, 9], so ∆x = 2 and we get
f (1) + f (3) f (3) + f (5) f (5) + f (7) f (7) + f (9)
T (4) = 2
+
+
+
2
2
2
2
= 2 · 124 = 248