A possible solution for Quiz 1
R
1.(10pts) Find sin (ln x)dx. (Hint: Use integration by parts)
We use integration by parts. Let u = sin(ln x) and dv = dx. Then we get
cos(ln x)
du =
dx and v = x. So we have,
x
Z
Z
sin (ln x)dx = x sin(ln x) − cos(ln x)dx.
(1)
We again use the similar integration by parts for the last integral. Then we get
Z
Z
cos(ln x)dx = x cos(ln x) + sin(ln x)dx.
(2)
Plugging (2) into (1) and with some algebra, we get
Z
x(sin(ln x) − cos(ln x))
+ C,
sin (ln x)dx =
2
where C is an arbitrary constant.
2.(10pts) Compute
R 1/√2
0
2x sin−1 x2 dx.
By the change of variable t = x2 , which gives dt = 2xdx, we get
Z 1/√2
Z 1/2
−1 2
2x sin x dx =
sin−1 tdt.
0
0
Be cautious of the upper and lower limits of the new integral!
√
√
x = 0 → t = 02 = 0 & x = 1/ 2 → t = (1/ 2)2 = 1/2
Or, you can also choose to calculate the indefinite integral
Z
F (x) = 2x sin−1 x2 dx
first, and then calculate the definite integral by using the Fundamental Theorem of
Calculus, i.e.
Z 1/√2
√
2x sin−1 x2 dx = F (1/ 2) − F (0)
0
Now we proceed by using integration by parts for definite integral, letting u =
1
sin−1 t and dv = dt. Then we get du = √
dt and v = t, so
1 − t2
Z 1/2
Z 1/2
t
−1
−1 1/2
√
sin tdt = t sin t 0 −
dt
introducing u = 1 − t2 gives
2
1
−
t
0
0
√
i1/2
1π √
π
3
=
+ 1 − t2
=
+
− 1.
26
12
2
0
You may also use the change of the variable method first to integrate sin−1 t. See page
443 of the textbook for more information about integrating general inverse functions.
3.(10pts) Find the volume of the solid generated by revolving the region of the xyplane bounded by the x-axis, y = ln x, y = 0, and x = e, about x-axis.
By using the disc method, we get the volume V of the solid as follows.
Z e
V =π
(ln x)2 dx.
1
We can continue by introducing the new variable u = ln x or using integration by
parts. Let me do the latter.
Let u = (ln x)2 and dv = dx. Then we have du = 2 lnx x dx and v = x.
Z e
Z e
2
2 e
ln xdx
(ln x) dx = π x(ln x) 1 − 2
π
1
1
= π (e − 2x(ln x − 1)]e1 ) = π(e − 2),
where we have used
R
ln xdx = x(ln x − 1) + C.
We can also use the shell method. In this case, we have
Shell Height = e − ey
Shell Radius = y,
so we get the volume of the solid
Z
V = 2π
1
y(e − ey )dy,
0
which gives the same answer π(e − 2).
2