Problem 1 (Exam 1 Fall 2015 - 4). The integral
of the following... [some options are given]
R
dx
(3−2x−x2 )3/2
can be transformed into which
solution. Complete the square downstairs:
Z
Z
Z
dx
dx
dx
=
=
(3 − 2x − x2 )3/2
(3 + 1 − 1 − 2x − x2 )3/2
(4 − (1 + x)2 )3/2
so we see that the useful substitution is (1 + x) = 2 sin(θ), dx = 2 cos(θ)dθ. The integral
becomes
Z
Z
Z
Z
dx
cos θ
1
2
1
1
dθ =
= 3/2
dθ =
sec2 (θ)dθ.
2
3/2
2
3/2
2
(4 − (1 + x) )
4
(cos (θ))
4
cos (θ)
4
Problem 2 (Exam 1 Fall 2015 - 9). Evaluate
R2
1
x3 ln(8x) dx.
solution. Integrate by parts choosing u = ln(8x), dv = x3 dx.
Z 2 4
Z 2
x
x4 2
3
x ln(8x) dx = ln(8x) · −
dx
4 1
1 4x
1
Z 2 3
x4 2
x
= ln(8x) · −
dx
4 1
1 4
x4 2 x4 2
= ln(8x) · − 4 1 16 1 1
16
1
16
−
−
= ln(16) − ln(8)
4
4
16 16
1
1
4
3
= 4 ln(2 ) − ln(2 ) − 1 −
4
16
15
3
= 16 ln(2) − ln(2) −
16
4
3
15
= 16 −
ln(2) −
4
16
61
15
=
ln(2) −
4
16