PRACTICE PROBLEMS: SET 5
MATH 101: PROF. DRAGOS GHIOCA
1. Problems
Problem 1. Compute
√
3
∫
arctan
1
Problem 2. Compute
( )
1
dx .
x
∫
cos(x) ln(sin(x)) dx .
Problem 3. Evaluate
∫
sec8 (x) dx.
1
2
MATH 101: PROF. DRAGOS GHIOCA
2. Solutions
Problem 1. We first evaluate the antiderivative and at the end we will plug in
the bounds of the integration.
( 1 )We will use integration by parts in order to evaluate
∫
the antiderivative arctan
( 1 ) x dx.
We use u = arctan x and dv = 1 · dx. Hence
du =
− x12
−1
dx .
( 1 )2 dx = 2
x
+1
1+ x
Also, v = x. Thus, integration by parts gives
( )
∫
1
arctan
dx
x
( ) ∫
1
−x
dx
−
x
x2 + 1
( ) ∫
x
1
+
dx .
= x arctan
x
x2 + 1
= x arctan
We evaluate separately the second integral:
∫
x
dx .
x2 + 1
We use the substitution u = x2 + 1, which gives du = 2x dx. Hence x dx =
So,
∫
∫ 1
∫
x dx
1
1
1
2 du
=
=
du = ln | u | + C .
x2 + 1
u
2
u
2
1
2
du.
We conclude
∫
( )
( )
1
1
1
arctan
dx = x arctan
+ ln | x2 + 1 | + C .
x
x
2
Now we evaluate the definite integral and obtain:
∫
√
3
arctan
1
( )
1
dx
x
(
( )
) √
1
1
x arctan
+ ln | x2 + 1 | |1 3
x
2
√
(2 3 − 3)π + 6 ln(2)
.
=
12
=
Problem 2. In order to integrate by parts, we first need to make a substitution.
We let u = sin(x). Then du = cos(x) dx. So,
∫
∫
ln(sin(x)) cos(x) dx = ln(u) du .
PRACTICE PROBLEMS: SET 5
3
Then we evaluate the integral of the logarithm using integration by parts:
∫
ln(u) · 1 du
∫
1
= ln(u) · u −
· u du
u
∫
= u ln(u) − 1 du
= u ln(u) − u + C .
We conclude that
∫
cos(x) ln(sin(x)) dx = sin(x) ln(sin(x)) − sin(x) + C .
Problem 3. First we manipulate algebraically, using the trigonometric identities
and get
sec8 (x) = (sec2 (x))3 · sec2 (x) = (1 + tan2 (x))3 sec2 (x).
Now we can use the substitution u = tan(x) because du = sec2 (x) dx. So,
∫
sec8 (x) dx
∫
=
(1 + u2 )3 du
∫
=
(1 + 3u2 + 3u4 + u6 ) du
u7
3u5
+
+C
5
7
3 tan5 (x) tan7 (x)
tan(x) + tan3 (x) +
+
+C.
5
7
= u + u3 +
=