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Answers to Practice Problems Exam 3
- a)
€
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.
Since x = 1 sin θ , then
x
1
= sin θ =
opposite
hypot.
. Label triangle with x = opposite
and 1 = hypot. The missing side (adjacent) will have length 1 − x
2
€
Thus, tan θ =
opposite
adjacent
=
x
1 − x
2
, so ( 1 − x
2
)
−
3
2
∫
dx =
x
1 − x
2
b)
€
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
c)
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
+ C =
1
4
sin
4
( x ) −
1
6
sin
6
( x ) + C.
d)
tan x csc xdx
∫
sin( x )
cos( x )
∫
sin( x )
dx =
cos( x )
∫
dx = sec( x )
∫
dx = ln sec( x ) + tan( x ) + C.
e)
(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
= x + 2 (sin x )(cos x )
∫
dx ( Now let u = sin x , so du = cos xdx. )
= x + 2 udu
∫
= x + 2
1
2
u
2
+ C = x + sin
2
( x ) + C.
f)
€
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
cos( u
) du
[ ]
(using a reduction formula)
1
3
cos
2
( u )(sin u )
3
2
3
sin( u )
[ ]
+ C =
1
3
cos
2
( 3 x )(sin( 3 x ))
3
2
3
sin( 3 x )
[ ]
+ C.
g)
sin x sin( 2 x ) dx
= sin x ( 2 (sin x )(cos x )) dx = 2 sin
2
( x )(cos x ) dx
= 2 u
2
du =
2
3
u
3
+ C =
2
3
sin
3
( x ) + C.
1. h)
cos
4
xdx
cos
3
( x )(sin x )
4
3
4
cos
2
( x
) dx
[ ]
(using a reduction formula)
cos
3
( x )(sin x )
4
3
4
1
2
( 1 + cos( 2 x )
) dx + C =
cos
3
( x )(sin x )
4
3
8
( x +
1
2
sin( 2 x )) + C.
(or :
cos
3
( x )(sin x )
4
3
8
x +
3
8
1
2
sin( 2 x )) + C =
cos
3
( x )(sin x )
4
3
8
x +
3
8
1
2
( 2 )(sin x )(cos x )) + C
cos
3
( x )(sin x )
4
3
8
x +
3
8
(sin x )(cos x ) + C .)
i)
For
2 x − 3
x ( x − 3 )( x
2
dx , use partial fractions :
2 x − 3
x ( x − 3 )( x
2
A
x
B
x − 3
Cx + D
x
2
€
Then 2 x − 3 = A ( x − 3 )( x
2
2
- 1 ) + ( Cx + D ) x ( x − 3 ).
For x = 3 : 3 = A ( 0 ) + B ( 3 )( 10 ) + ( Cx + D )( 0 ), so B =
1
10
For x = 1 : − 1 = A (− 2 )( 2 ) + B ( 1 )( 2 ) + ( C + D )( 1 )(− 2 ),
so − 1 = ( 1 )(− 4 ) + (
1
10
)( 2 ) + ( C + D )(− 2 ) so C +D = −
7
5
For x = 2 : 1 = A (− 1 )( 5 ) + B ( 2 )( 5 ) + ( 2 C + D )( 2 )(− 1 ),
so 1 = ( 1 )(− 5 ) + (
1
10
)( 10 ) + ( 2 C + D )(− 2 ) so 2 C +D = −
5
2
Since C + D = −
7
5
, then - C - D =
7
5
. Using this equation
together with 2 C +D = −
5
2
(add), we find that C = −
11
10
and since C +D = −
7
5
, then D = −
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.
j)
For
x
4
− x
2
dx , use partial fractions :
x
4
− x
2
x
2
( x
2
x
2
( x + 1 )( x − 1 )
A
x
B
x
2
C
x + 1
D
x − 1
€
Thus,
dx
x − 2
3
1
29
∫
= −
3
2
27
2
=
24
2
= 12.
dx
x − 2
3
1
29
∫
converges to 12.
⎛
⎝
⎜
⎞
⎠
⎟
c)
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.
d)
€
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
.
3. e)
€
dx
x + e
x
0
∞
∫
(Since we can't find the antiderivative of f(x) =
1
x + e
x
,
we will use the Comparison Test for Nonnegative Integrals.)
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
Since
dx
x + e
x
0
∞
∫
≤ e
− x
dx
0
∞
∫
and e
− x
dx
0
∞
∫
converges to 1,
then by the Comparison Test for Nonnegative Integrals,
dx
x + e
x
0
∞
∫
also converges.
f)
€
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.
g)
€
dx
x
4
2
∞
∫
(Since we can't find the antiderivative of f(x) =
1
x
4
,
we will use the Comparison Test for Nonnegative Integrals.)
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
