Download MTH 240 Lab,Winter 2008 Homework 3/Quiz 3 Assigned Homework

MTH 240 Lab,Winter 2008
Homework 3/Quiz 3
Jan. 29, 2008
Assigned Homework
Section 7.4: #12, 16, 19, 38
Section 7.5: #13, 23, 28, 38, 53, 62
Please grade: Section 7.4, problem 12.
Find
Z
0
1
x2
x−1
dx
+ 3x + 2
Solution
We use partial fractions; the denominator factors as x2 + 3x + 2 = (x + 1)(x + 2)
A
B
(A + B)x + (2A + B)
x−1
=
+
=
x2 + 3x + 2
x+1 x+2
(x + 1)(x + 2)
Equating coefficients on the left and right hand sides of the above equation we get A+B = 1 and 2A+B = −1.
Solving, A = −2 and B = 3. So
Z
1
x−1
dx
2
0 Z x + 3x + 2 Z
1
1
−2
3
=
dx +
dx
x
+
1
x
+
2
0
0
= −2 ln(x + 1) + 3 ln(x + 2)|10
3
= −2 ln 2 + 3 ln
2
27
= ln( )
32
Quiz
Z
x ln(x)
√
dx.
x2 − 1
1. Find
π
3
Z
2. Find
π
4
ln(tan(x))
dx.
sin(x) cos(x)
Solution
p
1. Let tZ= x2 − 1. Then
x ln(x)
√
I=
dx
2
Z x −1
1
=
ln(t2 + 1) dt
2
Use integration by part
with u = ln(t2 + 1), dv = dt. Then
Z
1
t2
I = t ln(t2 + 1) −
dt
2
t2 + 1
1
= t ln(t2 + 1) − t + arctan(t) + C
2
p
p
p
= x2 − 1 ln(x) − x2 − 1 + arctan( x2 − 1) + C
2. Let u = tan(x). Then
Z π3
ln(tan(x))
dx
π
sin(x) cos(x)
4
Z π3
ln(tan(x))
=
sec2 (x) dx
π
tan(x)
4
Z √3
ln(u)
du
=
u
1
1
= (ln 3)2 .
8