Math 222 Worksheet 2 (1/30 - 2/3)
The problems on this worksheet are intended to be done in groups. Aim for about ten minutes per
problem; if you find yourself using more, move on to the next one. Don’t worry about finishing all of the
problems, but do try to have at least started each one before the end of class.
1. Compute
the following Rintegrals. Integration byR parts may be useful.
R
2
(a) x ln 2xdx
(b) x cos 7xdx
(c) x3 ex dx
(a)R Using integration by
we’ll take u = lnR2x andR dv = xdx. du is then x1 dx, and v is 12 x2 . So
R 1parts,
1 2
21
uv − vdu = 2 x ln 2x − 2 x x dx = 12 x2 ln 2x − 12 xdx. xdx = 21 x2 + c, so inserting this in we obtain
our final answer 12 x2 ln 2x − 14 x2 + c.
(b) Again, we’ll
byR parts. We’ll take u = x, dv = cos 7xdx. So du = dx and v = 71 sin 7x.
R use integration
1
1
So we have uv − vdu = 7 x sin 7x − 17 sin 7xdx = 17 x sin 7x + 49
cos 7x + c, which is our final answer.
(c) For this one, we’ll start with a substitution. Since we may do integration by parts later, we’ll use the
R
R
2
variable w instead Rof u. We take w = x2 . Then dw = 2xdx, so we have x3 ex dx = 12 wew dw. So now we
just need to know R wew dw. For this, Rwe’ll use integration by parts. Take u = w, dv = ew dw. So du = dw,
v = ew . Therefore wew dw = wew − ew dw = wew − ew + c. Recalling that w = x2 and that our original
2
2
integral has a 21 , we now have 12 x2 ex − 21 ex + c as our final answer.
2. Compute
the following integrals.
R
R
(a) x sec2 xdx
(b) x(ln x)2 dx
(c)
R
e3x sin 2xdx
R
R
(a) Take uR = x, dv = sec2 xdx. Then du = dx and v = tan x. So we have uv − vdu = x tan x−
tan xdx.
R
sin x
ToR compute tan xdx, rewrite tan x as cos
and
substitute
u
=
cos
x.
Then
du
=
−
sin
xdx,
so
tan
xdx =
x
− u1 du = − ln |u| + c = − ln | cos x| + c. Inserting this into the result of our integration by parts, we have
our final answer x tan x + ln | cos x| + c.
R
1 2
x
we have uv − vdu = 21 x2 (ln x)2 −
(b) Take u = (ln x)2 , dv = xdx. Then du = 2 ln
x dx and v = 2 x . So
R
R
1 2 2 ln x
1 2
2
2x
x dx
R = 2 x (ln x) − x ln xdx. Now we have a new question: x ln xdx.
To find x ln xdx,
by
with u = ln x and dv = xdx. So du = x1 dx and
R we do integration
R 1parts again,
1 2
1 2
1 2
v = 2 x . We have x ln xdx = 2 x ln x − 2 xdx = 2 x ln x − 14 x2 + c.
R
Inserting this back into our original expression, we have our final answer x(ln x)2 dx = 21 x2 (ln x)2 −
1 2
1 2
2 x ln x + 4 x + c.
R
Note: The tabular method demonstrated in class would also have been appropriate here; since there were
only two steps, it’s a bit of a judgment call.
R
(c) Take u = e3x , dv = sin 2xdx. Then du = 3e3x and v = − 21 cos 2x. So uv − vdu = − 21 e3x cos 2x +
R
3
e3x cos 2xdx.R
2
To compute e3x cos 2xdx, we’ll do Rintegration by parts again,
with u = e3x and dv = cos 2xdx. Then
R
du = 3e3x dx and v = 12 sin 2x. So uv − vdu =R 12 e3x sin 2x − 32 e3x sin 2xdx.
R
Putting theseR two results together, we have e3x sin 2xdx = − 12 e3x cos 2x + 34 e3x sin 2x − 94 e3x sin 2xdx.
If we write I = e3x sin 2xdx, then this says I = − 21 e3x cos 2x + 34 e3x sin 2x − 94 I + c. Now we can solve for
I.
1 3x
3 3x
Adding 94 I to both sides, we have 13
cos 2x
Dividing by 13
4 I = −2e
4 , we have
R +3x4 e sin 2x + c.2 3x
2 3x
3 3x
3 3x
I = − 13 e cos 2x+ 13 e cos 2x+c. So our final answer is e sin 2xdx = − 13 e cos 2x+ 13
e cos 2x+c.
3. Find a reduction formula for In =
R
xn e3x dx, and use it to calculate I5 .
R
R
Take u = xn , dv = e3x dx. Then du = nxn−1 dx, v = 13 e3x . The uv − vdu = 31 xn e3x − 13 nxn−1 e3x dx.
Note that n is justR a constant, as far as theR integral is concerned; so the n can come out of the integral,
and we now have xn e3x dx = 13 xn e3x − n3 xn−1 e3x dx. This rightmost integral is just In−1 , so we have
In = 31 xn e3x − n3 In−1 .
R
This is almost a complete reduction formula - we just need a base case. n = 0 is easiest: I0 = e3x dx =
1 3x
+ c.
3e
So our final answer is I0 = 13 e3x + c, In = 31 xn e3x − n3 In−1 .
R
4. Let In = xn cos(3x)dx. A reduction formula for this integral is In = 13 xn sin 3x + n9 xn−1 cos 3x −
R 5
n2 −n
x cos 3xdx.
9 In−2 . Using this formula, find
x5 cos 3xdx is I5 . The reduction formula says that I5 = 13 x5 sin 3x + 95Rx4 cos 3x − 20
9 I3 . Applying it
again to n = 3, we have I3 = 31 x3 sin 3x + 39 x2 cos 3x − 96 I1 . I1 is the integral x cos 3xdx. Using integration
by parts with u = x, dv = cos 3xdx, we get I1 = 31 x sin 3x + 19 cos 3x + c. Inserting this into our expression
for I3 , we have:
R
1 3
1
2
2
x sin 3x + x2 cos 3x − x sin 3x −
cos 3x + c
3
3
9
27
And inserting this into our expression for I5 , we get our final answer:
I3 =
I5 =
1 5
5
20
20
40
40
x sin 3x + x4 cos 3x − x3 sin 3x − x2 cos 3x + x sin 3x +
cos 3x + c
3
9
27
27
81
243
5. Compute
the followingRintegrals. The technique of partial
fractions will be useful.
R
R
x+1
cos x
dx
(b) (x+2)(x+3)(x−1)
dx
(c) sin2 x+4
(a) x2 −3x+2
dx
sin x+3
(a) x2 − 3x + 2 = (x − 2)(x − 1). We can rewrite
Multiplying both sides by x − 1,
1
we have x−2
A(x−2)
1
x−1 = x−1
Multiplying instead by x − 2, we have
1
1
1
x2 −3x+2 = x−2 − x−1 .
R 1
R
1
dx =
Therefore, x2 −3x+2
x−2 −
1
x−1
1
x2 −3x+2
= A+
A
B
= x−1
+ x−2
; we just
B(x−1)
x−2 . Setting x =
need to find A and B.
1, this says A = −1.
+ B. Setting x = 2, this says B = 1. So we have that
dx = ln |x − 2| − ln |x − 1| + c, our final answer.
x+1
A
B
C
(b) We can rewrite (x+2)(x+3)(x−1)
as x+2
+ x+3
+ x−1
.
−1
To find A: Multiply both sides by x + 2, and set x = −2. We get 1·−3
= A, so A = 31 .
−2
= B, so B = − 21 .
To find B: Multiply both sides by x + 3, and set x = −3. We get −1·−4
2
To find C: Multiply both sides by x − 1, and set x = 1. We get 3·4 = C, so C = 61 .
1/3
1/2
1/6
x+1
= x+2
− x+3
+ x−1
. Integrating these gives our final answer 31 ln |x + 2| −
So (x+2)(x+3)(x−1)
1
3| + 6 ln |x − 1| + c.
1
2
ln |x +
(c) First of all, this isn’t a partial fractions problem yet - there’s too many trig Rfunctions. We’ll get
cos x
dx =
rid of them with a substitution, taking u = sin x. Then du = cos xdx, and we have sin2 x+4
sin x+3
R
1
du.
u2 +4u+3
1
A
B
u2 + 4u + 3 factors as (u + 1)(u + 3). We can rewrite u2 +4u+3
as u+1
+ u+3
; using the Heaviside
R
R
1/2
1/2
1
1
1
trick, we get that A = 2 and B = − 2 . So u2 +4u+3 du =
u+1 − u+3 du. Integrating, we have
1
2
ln |u + 1| −
1
2
ln |u + 3| + c. Plugging u = sin x back in, we get our final answer
1
1
ln | sin x + 1| − ln | sin x + 3| + c
2
2