Calculus II
Practice Exam 2, Answers
Find the indefinite integrals of the following functions.
x
arccos
dx
1 x
Answer. Let u arccosx du dx 1 arccosx
dx 1 x
1.
2
x2 . Then
2
2.
ln x x 1
C
u2 C arccosx
2
2
udu
2
Answer. Let u
dx
ln x
1 du dx x. We get
ln x x 1 dx 2
3.
2
u2 du
ln x 3 1 C 3
cos3 x sin2 xdx
1 sin x:
cos x sin xdx 1 sin x sin x cos xdx 2
Answer. Here we use the famous trig identity: cos 2 x
3
Now, let u
sin x du
2
2
2
cos xdx:
1 u u du u u du 2
4.
2
2
sin3 x
3
4
sin5 x
5
C
dx
x2 x 1
Answer. We need the partial fraction expansion
A
x 1
x2 x 1
B
x2
C
x 1 Putting the right hand side over a common denominator, we get this equality of the numerators: 1
1 B x 1 Cx2
x 0: 1
B so that B
1
x
2
coefficient of x : 0
Thus
dx
x2 x 1
dx
x
dx
x2
1: 1
A C C
so that A
1 dx
x
lnx
1 1
x ln x 1 C 2
Ax x
5.
x x dx xx
3 2
1 dx
1 2
2 5
x
5
2
2 3
x
3
2
C
This is here to remind you to try the easy things first; don’t just assume that if an integral appears in this
section, it will need some complicated substitution or integration by parts.
5
Answer. Again, we seek the partial fractions decomposition
6.
x x2
dx
4x
A
Bx C
x
5 x x 4x 5
This leads to the equation of numerators: 1 A x
4x 5 Bx Cx.
1
x 0 : 1 5A so that A 5
4
coefficient of x : 0 4A C so that C 5
1
coefficient of x : 0 A B so that B 5
1
4x
x2
2
2
2
2
This gets us to
dx 5
To integrate the last term we use a little algebra:
x x2
dx
4x
1
5
dx
x
4x4
x
x2
5
dx
x 2 dx
2dx
5
x 2 1 x 2 1
12 ln x 2 1 2 arctan x 2 C 15 ln x 12 ln x 2 1 2 arctan x 2 C 5
4x4
x
x2
dx
2
2
2
So, finally
dx
4x
x x2
7.
2
x2 sin xdx
cosx:
x sin xdx x cosx 2 x cos xdx Another integration by parts handles the last integral: u x dv cos xdx du dx v x cos xdx x sin x sin xdx x sin x cosx Answer. Let u
x dv 2
sin xdx du
2xdx v
2
finally giving
x2 sin xdx
Calculate the definite integral.
2
x2 cos x
2 x sin x cos x
C
sin x:
2
8.
0
x 3x 1 2x 3 dx
2
2
x 3x 1 du 2x x 3x 1 2x 3 dx Answer. Here the substitution u
2
0
e
9.
2
2
2
u2 du
1
u3
3
9
35
1
1
3
ln 2x dv x dx du dx x v x 3:
x
1
x
x
x ln 2x dx x dx ln 2x ln 2x 3 3
3 9
2
Answer. Integrate by parts with u
Then
e
x2
dx
4x
5
2
3
0
1
dx
x 2
2
3
2
x
x
3 ln 2x 9 x2 ln 2x dx
1
2
3
3
2
0
9
x2 ln 2x dx
1
10.
3 works:
3
e
1
arctan x
2
2
0
e3
ln 2e
3
arctan4
3
e3
9
1
9 1
ln 2
3
arctan2
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