Z
π/4
x5 sin(x3 ) dx
0
The version of this problem that I gave in class was way too difficult. I hope
you got my email and didn’t spend too much time on this problem. However,
this one is doable.
To do this problem, you need to use integration by parts.
u = x3
dv
= x2 sin(x3 )
dx
du
= 3x2
dx
v = − cos(x3 )/3
This gives
π/4
−x3 cos(x3 )/30 −
Z
π/4
3x2 (− cos(x3 )/3) dx
Z π/4
x2 cos(x3 ) dx
= −(π/4)3 cos (π/4)3 /3 +
0
0
3
3
= −(π/4) cos (π/4)
/3 + sin(x3 )/3
π/4
0
= −(π/4)3 cos (π/4)3 /3 + sin( π/4)3 /3.
For this second integral, you should also use integration by parts.
Z
e5
x log x dx
1
We select:
u = log x
dv
=x
dx
du
1
=
dx
x
x2
v=
2
1
Then
Z
e5
x log x dx
1
e5 Z e5
x2
x
log x −
dx
=
2
2
1
1
2 e5
5e10
x
=
−
2
4 1
1
10
e 0 1
5e
−
+
=
2
4
4
10
1
9e
+
=
4
4
Now for the third integral:
Z
π/3
sec3 (x) dx
0
This integral can be done by integration by parts, selecting u = sec x and
dv
2
dx = sec x. However, I’ll instead write this integral as
Z
0
which is the same as
Z
0
π/3
π/3
cos(x)
dx
cos4 (x)
cos(x)
dx
(1 − sin2 (x))2
We will make the substitution u = sin(x), and cancel out
numerator:
Z √3/2
1
du.
(1 − u2 )2
0
du
dx
= cos(x) with the
This problem is a little “unfair” in that it requires a partial fraction decomposition with repeated linear factors; we expect the integrand to look like
C
B
D
A
+
+
+
u − 1 (u − 1)2
u + 1 (u + 1)2
for an appropriate A, B, C, and D. Setting this expression equal to 1 and
multiplying through by (u2 − 1)2 gives
1 = A(u + 1)2 (u − 1) + B(u + 1)2 + C(u − 1)2 (u + 1) + D(u − 1)2 .
Considering the u3 term gives A + C = 0, or C = −A. Picking u = 1 gives
B − 41 . Picking u = −1 gives D = 14 . Picking u = 0 gives 1 = −A + B + C + D,
2
but −A = C, so this comes out to 2C =
the integral is
1
4
Z
√
3/2
0
1
2
or C = 14 . Therefore, A = − 14 , and
1
1
1
1
+
du
+
+
1 − u (u − 1)2
u + 1 (u + 1)2
And we integrate to get
√
√
1
1
1
+1 .
− log(1 − 3/2) − √
+ 1 + log(1 + 3/2) − √
4
( 3/2 − 1)2
( 3/2 − 1)2
I didn’t check my work on this integral, so the value might not be right. Do it
yourself and see if you get the same answer!
Z
π/6
sin2 (x) cos2 (x) dx
0
We want to use the double-angle formulas to reduce this integral
Z
π/6
0
1 − cos(2x)
2
1 + cos(2x)
2
.
The cross-terms cancel out, leaving us with
1
4
Z
π/6
1
4
Z
0
which can be writen as
1 − cos2 (2x) dx
π/6
sin2 (2x) dx
0
We will again use the double-angle formula:
1
4
Z
0
π/6
1 cos(4x)
−
dx
2
2
π/6
1
π
, and the second term comes out to − 32
sin(4x)0 ,
The first term comes out to 48
√
which is − 3/64.
For the last integral, the trick is to make a trigonometric substitution x =
3
tan t.
2
1
√
dx.
1 + x2
1
Z arctan(2)
sec2 t
p
=
dt
tan2 t 1 + tan2 (t)
π/4
Z arctan(2)
sec2 t
=
tan2 t sec t
π/4
Z arctan(2)
sec t
=
tan2 t
π/4
Z arctan(2)
cos t
=
2
sin
t
π/4
arctan(2)
−1 =
sin t π/4
Z
x2
We can compute sin(arctan(2)) = √25 by drawing out a right triangle with side
√
lengths 1, 2, and 5. This gives us
√
5 √
+ 2
−
2
4