Solution to Quiz 6
• (50 pts)
Z
e3x sin(2x)dx
We start with u = e3x and dv = sin(2x)dx.
R
Then we have du = 3e3x dx (use chain rule here) and v = sin(2x)dx =
− cos(2x)
(note that (cos(2x))0 = − sin(2x) · 2 and (− cos(2x)
)0 = sin(2x).
2
2
Using integration by parts, we have
R 3x
R
cos(2x)
cos(2x)
3x
3x
e
sin(2x)dx
=
e
·
(−
)
) · 3e
−
(−
|{z} | {z } |{z}
{zdx}
|
|
|
{z2 }
{z2 }
u
dv
u
du
v
v
R
3x
= − e cos(2x)
+ 32 e3x cos(2x)dx
2
R
R
3x
Now we have e3x sin(2x)dx = − e cos(2x)
+ 23 e3x cos(2x)dx. We do not
2R
get the answer now. We need to find e3x cos(2x)dx using IBP again.
Let u = e3x and dv = cos(2x)dx. Then we have du = 3e3x dx (use chain
R
(note that (sin(2x))0 = cos(2x) · 2
rule here) and v = cos(2x)dx = sin(2x)
2
and ( sin(2x)
)0 = cos(2x).
2
Using integration by parts, we have
R 3x
sin(2x) R sin(2x)
3x
3x
e
cos(2x)dx
−
· 3e
=
e
·
|
|{z} | {z } |{z}
{zdx}
2 }
2 }
| {z
| {z
u
e3x sin(2x)
2
dv
u
du
v
v
R
3
− 2 e3x sin(2x)dx
R
R
3x
Now we have e3x cos(2x)dx = e sin(2x)
− 32 e3x sin(2x)dx.
2
R 3x
R
3x
3
+
e cos(2x)dx and
Now we combine e3x sin(2x)dx = − e cos(2x)
2
2
R 3x
R 3x
e3x sin(2x)
3
e cos(2x)dx =
− 2 e sin(2x)dx to get
2
Z
Z
e3x cos(2x) 3 e3x sin(2x) 3
3x
e sin(2x)dx = −
+
−
e3x sin(2x)dx
2
2
2
2
Z
3x
3x
e cos(2x) 3 e sin(2x) 3 3
(0.0.1)
=−
+ ·
− ·
e3x sin(2x)dx
2
2
2
2 2
Z
e3x cos(2x) 3e3x sin(2x) 9
+
−
e3x sin(2x)dx
=−
2
4
4
Thus
Z
Z
9
−2e3x cos(2x) + 3e3x sin(2x)
3x
e3x sin(2x)dx =
+ c,
e sin(2x)dx +
4
4
R
3x
3x sin(2x)
(1 + 49 )( e3x sin(2x)dx) = −2e cos(2x)+3e
+ c,
4
R 3x
−2e3x cos(2x)+3e3x sin(2x)
13
( e sin(2x)dx) =
+ c and
4
R4 3x
−2e3x cos(2x)+3e3x sin(2x)
4 −2e3x cos(2x)+3e3x sin(2x)
e sin(2x)dx = 13
+
C
=
+ C.
4
13
=
MATH 1760 : page 1 of 3
Solution to Quiz 6
MATH 1760 page 2 of 3
• (50 pts)
x4 − 3x3 − x2 + 10x − 10
dx
x2 − 3x + 2
Solution: From long division
Z
F IGURE 1. Graph for problem 6
, we have x4 − 3x3 − x2 + 10x − 10 = (x2 − 3)(x2 − 3x + 2) + x − 4. So
2
2 −3x+2)+x−4
x−4
x−4
= (x −3)(x
= x2 − 3 + x2 −3x+2
For x2 −3x+2
, we can
x2 −3x+2
x2 −3x+2
x−4
x−4
A
B
try the partial fraction x2 −3x+2 = (x−1)(x−2) = x−1 + x−2 . Multiplying
(x − 1)(x − 2), we have x − 4 = A(x − 1) + B(x − 2). Plugging x = 2
x−4
2
3
and x = 1, we get A = −2 and B = 3. Thus x2 −3x+2
= − x−1
+ x−2
R
4
3
2
4
3
2
+10x−10
2
3
+10x−10
and x −3xx2−x
= x2 − 3 − x−1
+ x−2
. Hence x −3xx2−x
dx =
−3x+2
−3x+2
R
3
3
2
x2 − 3 − x−1
+ x−2
dx = x3 − 3x − 2 ln |x − 1| + 3 ln |x − 2| + C.
x4 −3x3 −x2 +10x−10
Solution to Quiz 6
MATH 1760 page 3 of 3
• (10 pts)
Z
Bonus problem
arcsin(3x)dx
d
Solution: Let u = arcsin(3x) and dv = dx. Recall that dx
(arcsin(ax)) =
a
√
.
1−a2 x2
R
3
dx
and
v
=
dx = x. Using integration by parts, we
Then du = √1−9x
2
R
R
R 3x
3
√
have arcsin(3x)dx = arcsin(3x)x− x· √1−9x
dx.
2 dx = arcsin(3x)x−
1−9x2
du
2
We can use the substitution u = 1 − 9x , du = −18xdx and − 18 = xdx to
find
√
√
R 3x
R 1
R 3
u
1−9x2
√
√ · (− du ) = −
√ du = −
+
C
=
−
+ C. Thus
dx
=
2
18
3
3
u
6
u
1−9x
√
R
2
arcsin(3x)dx = x arcsin(3x) + 1−9x
+ C.
3