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Solutions to practice problems on integration and series, including using techniques such as partial fractions, reduction formulas, and the comparison test. Topics covered include improper integrals, definite integrals, and infinite series.
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Let x = 1 sin θ , so dx = cos( θ ) dθ. Then ( 1 − x
2
)
−
3
2
∫
dx =
cos( θ )
( 1 − sin
2
( θ ))
3
2
∫
dθ
€
=
cos( θ )
(cos
2
( θ ))
3
2
∫
dθ =
cos( θ )
cos
3
( θ )
∫
dθ =
1
cos
2
( θ )
∫
dθ = sec
2
( θ )
∫
dθ = tan( θ ) + C.
x
1
opposite
hypot.
2
€
Thus, tan θ =
opposite
adjacent
=
x
1 − x
2
, so ( 1 − x
2
)
−
3
2
∫
dx =
x
1 − x
2
€
Let x = 4 tan θ , so dx = 4 sec
2
( θ ) dθ. Then
dx
x
2
∫
=
4 sec
2
( θ )
( 16 tan
2
( θ ) + 16 )
1
2
∫
dθ
€
=
4 sec
2
( θ )
( 16 (tan
2
( θ ) + 1 ))
1
2
∫
dθ =
4 sec
2
( θ )
( 16 sec
2
( θ ))
1
2
∫
dθ =
4 sec
2
( θ )
4 sec( θ )
∫
dθ = sec( θ )
∫
dθ
€
=lnsec( θ ) + tan( θ ) + C. Since x = 4 tan θ , then
x
4
= tan θ =
opposite
adjacent
.
Label triangle with x = opposite and 4 = adjacent. The missing side
(hypot.) will have length x
2
hypot.
adjacent
=
x
2
4
and tan( θ ) =
x
4
, so
dx
x
2
∫
= ln
x
2
4
x
4
sin
3
x cos
3
∫
xdx = (sin
3
x )(cos
2
∫
x )(cos x ) dx = (sin
3
x )( 1 − sin
2
∫
x )(cos x ) dx
= ( u
3
)( 1 − u
2
∫
) du ( where u = sin( x ) and du = (cos x ) dx )
= u
3
− u
5
∫
du =
1
4
u
4
1
6
u
6
1
4
sin
4
( x ) −
1
6
sin
6
( x ) + C.
∫
∫
∫
∫
(cos x + sin x )
2
∫
dx = cos
2
( x ) + 2 (cos x )(sin x ) + sin
2
( x )
∫
dx
= (sin
2
( x )
∫
2
( x )) + 2 (cos x )(sin x ) dx = 1 + 2 (cos x )(sin x )
∫
dx
∫
∫
1
2
2
2
€
Let u = 3 x , so
1
3
( )
du = dx. Then cos
3
( 3 x
∫
) dx =
1
3
cos
3
( u
∫
) du
1
3
cos
2
( u )(sin u )
3
2
3
1
3
cos
2
( u )(sin u )
3
2
3
1
3
cos
2
( 3 x )(sin( 3 x ))
3
2
3
sin x sin( 2 x ) dx
= sin x ( 2 (sin x )(cos x )) dx = 2 sin
2
( x )(cos x ) dx
2
2
3
3
2
3
3
4
cos
3
( x )(sin x )
4
3
4
2
cos
3
( x )(sin x )
4
3
4
1
2
cos
3
( x )(sin x )
4
3
8
1
2
cos
3
( x )(sin x )
4
3
8
3
8
1
2
cos
3
( x )(sin x )
4
3
8
3
8
1
2
cos
3
( x )(sin x )
4
3
8
3
8
2
2
2
€
Then 2 x − 3 = A ( x − 3 )( x
2
2
1
10
1
10
7
5
1
10
5
2
7
5
7
5
5
2
11
10
7
5
3
10
€
Thus
2 x − 3
x ( x − 3 )( x
2
dx =
1
x
1
10
x − 3
−
11
10
x −
3
10
x
2
dx
= ln x +
1
10
ln x − 3 +
−
11
10
x
x
2
dx
−
3
10
x
2
dx
= ln x +
1
10
ln x − 3 −
11
10
1
2
( )
1
u
du
−
3
10
arctan( x ) + C
= ln x +
1
10
ln x − 3 −
11
20
ln x
2
3
10
arctan( x ) + C.
4
2
4
2
2
2
2
2
€
Thus,
dx
x − 2
3
1
29
∫
= −
3
2
27
2
=
24
2
= 12.
dx
x − 2
3
1
29
∫
converges to 12.
⎛
⎝
⎜
⎞
⎠
⎟
x
3
−∞
− 2
∫
dx = lim
t →−∞
2 x
− 1 / 3
x = t
− 2
∫
dx = lim
t →−∞
x = t
x
2 / 3
[ ]
= lim
t →−∞
2 / 3
−
t
2 / 3
€
=
3
2
(− 2 )
2 / 3
⎛
⎝
⎜
⎞
⎠
⎟
− ∞ = −∞, so
2
x
3
−∞
− 2
∫
dx diverges.
€
dx
1 − x
2
0
1
∫
= lim
t → 1
−
dx
1 − x
2
x = 0
t
∫
= lim
t → 1
− x = 0
t
arcsin( x ) [ ]
€
=lim
t → 1
−
(arcsin( t ) − arcsin( 0 )) = lim
t → 1
−
(arcsin( t ) − 0 ) =
π
.
€
dx
x + e
x
0
∞
∫
(Since we can't find the antiderivative of f(x) =
1
x + e
x
,
Thus, x + e
x
≥ 0 + e
x
, so x + e
x
≥ e
x
Hence,
x + e
x
e
x
so
dx
x + e
x
0
∞
∫
e
x
dx =
0
∞
∫
e
− x
dx
0
∞
∫
e
− x
dx
0
∞
∫
= lim
t →∞
e
− x
dx
x = 0
t
∫
= lim
t →∞
x = 0
t
− e
− x
[ ]
= lim
t →∞
x = 0
t
e
x [ ]
= lim
t →∞
e
t
e
0
= lim
t →∞
e
t
x
0
∞
∫
− x
0
∞
∫
− x
0
∞
∫
x
0
∞
∫
€
xdx
x
2
6
∞
∫
= lim
t →∞
xdx
x
2
x = 6
t
∫
= lim
t →∞
1
2
1
u
x = 6
t
∫
du
⎛
⎝
⎜
⎞
⎠
⎟
= lim
t →∞
x = 6
t
1
2
ln u
[ ]
€
=lim
t →∞
x = 6
t
1
2
ln x
2
[ ]
= lim
t →∞
1
2
ln t
2
1
2
ln( 38 )
( )
= ∞ −
1
2
ln( 38 ) = ∞,
so the integral
xdx
x
2
6
∞
∫
diverges.
€
dx
x
4
2
∞
∫
(Since we can't find the antiderivative of f(x) =
1
x
4
,
Since we're integrating from 2 to ∞, x ≥ 2. Thus, x ≥ 0 , so x
4
4
Therefore,
x
4
x
4
, so
dx
x
4
2
∞
∫
x
4
dx
2
∞
∫
Note that, since
x
4
dx
2
∞
∫
has the form
x
p
dx
a
∞
∫
, where p > 1,
then we know that
x
4
dx
2
∞
∫
converges.
Or, we can prove this using limits:
x
4
dx
2
∞
∫
= lim
t →∞
x
− 4
dx
x = 2
t
∫
= lim
t →∞ x = 2
t
−
1
3
x
− 3
[ ]
= lim
t →∞
x = 2
t
3 x
3 [ ]
= lim
t →∞
3 t
3