Problem 1 (L3 - 10).
Problem 2 (L3 - 29).
Problem 3 (L4 - 17).
Problem 4 (L4 - 44).
Problem 5 (W.A.1 - 8). Evaluate
R
2
x3 ex dx.
solution. Let z = x2 , dz = 2x dx. With a bit of algebra, the integral becomes
Z
Z
Z
1
3 x2
2 x2
zez dz.
x e dx = x e x dx =
2
Now run integration by parts once.
Problem 6 (W.A.1 - 10). Evaluate
R
x3 (9x2 )12 dx.
solution. Let z = 9x2 , dz = 18x dx. The integral becomes
Z
Z
Z
Z
1
3
2 12
2
2 12
12
x (9x ) dx = x (9x ) x dx = (z/9)(z) (dz/18) =
z 13 dz.
162
Now use the power rule.
Problem 7 (W.A.1 - 11). Evaluate
R
sin(ln 3x) dx.
solution. Let z = ln(3x), dz = dx/x. Observe that, by our substitution, x = ez /3; this is
needed to get the “left over” pieces of our integral in terms of z. The integral becomes
Z
Z
Z
Z
1
00
“
z
ez sin(z) dz.
sin(ln 3x) dx =
sin(z)xdz = sin(z)(e /3) dz =
3
I use the quotation marks as a reminder that the object in the middle with the mixture of
xs and zs makes zero sense to us as an integral. It is just a step that allows us to more
easily see what the substitution is doing. When making a substitution we must always get
the integrand in terms of a single variable before proceeding.
Now the integral on the far right hand side is one of the type we did in class; run integration
by parts twice to build an equation which allows us to solve for the integral as though it’s a
variable in an algebraic expression.
R
Problem 8 (W.A.1 - 19). Evaluate (sin−1 (x))2 dx.
1
2
solution. Let z = sin−1 (x), dz = √1−x
=
2 dx. By our substitution sin(z) = x, so 1 − x
2
2
1 − sin (z) = cos (z). The integral becomes
Z
Z
Z
z2
z2
p
dz =
dz = z 2 sec(z) dz.
2
cos(z)
1 − sin (z)
Now use integration by parts to kill the polynomial term. I suggest the tabular method for
this problem.
Problem 9 (L2 - N.Y.T.I. 3). Evaluate
Rt
0
ex sin(t − x) dx.
solution. You may choose to make the substitution z = t − x, dz = −dx, noting that this
forces x = t − z, so ex = et−z = et · e−z . Don’t forget to alter the limits of integration!
Since the integral is not taken with respect to t, the term et is a constant. Thus we have
Z t
Z 0
Z 0
Z t
−z
t
t
−z
t
x
e−z sin(z) dz
e sin(z) dz = −e
e · e sin(z) dz = e
e sin(t − x) dx =
0
t
t
0
which yields to integration by parts as in W.A.1 - 11 on the previous page.
You could also just integrate by parts right out of the gate. In either case, the trick is to be
careful with the limits.