1
Not all solutions are complete but hopefully, there is enough here that you can complete the
solution on your own. Please also see the list of topics for the second midterm for some useful
calculations.
(1)(a) Done in class.
(1)(b) Make the substitution u = cos(x)
(1)(c) This was a homework problem which we also did in class and in office hours today. Write
cot2 (x) = cscx (x) 1 and substitute to obtain
Z
Z
Z
5
3
csc (x)dx 2 csc (x)dx + csc(x)dx
Then reduce the power of cosecant by integration by parts. For instance, put f 0 =
g = csc3 (x). Then f = cot(x) and g 0 = 3csc3 (x)cot(x). This gives
Z
Z
5
3
csc (x)dx = csc (x)cot(x) 3 csc3 (x)cot2 (x)dx
csc2 (x) and
Again use the formula cot2 (x) = cscx (x) 1 to get
Z
Z
Z
5
3
5
csc (x)dx = csc (x)cot(x) 3 csc (x)dx + csc3 (x)dx
R
Now this is an equation in csc5 (x)dx, which we solve to get
✓
◆
Z
Z
1
5
3
3
csc (x)dx =
csc (x)cot(x) + csc (x)dx .
4
R
R
R dx
Repeat to compute the csc3 (x)dx in terms of csc(x)dx = sin(x)
.
(1)(d) Make the substitution u = cot(x).
(1)(g) The rational function is proper so we do not need to do Euclidean division. The partial
fractions expansion is
4x4
20x3 + 30x2 110x + 115
2
7
10
=
+ 2
2
2
2
(x 3) (2x + 5)
x 3 (x 3)
2x + 5
p
p
To integrate the quadratic term, make the substitution 2 x = 5 tan(t).
(1)(h) As above, the partial fractions expansion is
10x2 28x 156
=
(x + 3)(x + 5)(2x 3)
1
9
+
x+3 x+5
6
2x
x2 + 3
x2 + 2x + 3 2x
1
=
= 2
2
2
2
2
(x + 2x + 3)
(x + 2x + 3)
x + 2x + 3
(x2
3
(1)(i)
2x
+ 2x + 3)2
2
So
Z
Z
Z
(x2 + 3)dx
dx
2xdx
=
2
2
2
2
(x + 2x + 3)
x + 2x + 3
(x + 2x + 3)2
Complete the square to get x2 + 2x + 3 = (x + 1)2 + 2 and make the substitution u = x + 1 (du = dx)
to get
Z
Z
Z
Z
Z
Z
(x2 + 3)dx
du
2(u 1)du
du
2udu
2du
=
=
+
2
2
2
2
2
2
2
2
2
(x + 2x + 3)
u +2
(u + 2)
u +2
(u + 2)
(u + 2)2
For the middle integral
make the substitution v = u2 + 2. For the first and third integrals make the
p
substitution u = 2tan(t).