Part II: The Answers
488
Answers
701–800
or −
=
1
2
1
3
du x
dx. With these substitutions, you get the following:
−
( )
( )
= − −
( )
(
) +
=





 +
∫
1
2
1
2
1
2
1
2
csch
coth
csch
csch
u
u du
u
C
x
C
798.
7
12
The given problem is
cosh
cosh
ln
ln
x
x
dx
2
2
3
1
−
∫
To find the antiderivative, begin by using the hyperbolic identity cosh
2
x – 1 = sinh
2
x:
cosh
cosh
cosh
sinh
coth csch
ln
ln
ln
ln
ln
l
x
x
dx
x
x
dx
x
x
2
2
3
2
2
3
2
1
−
=
=
∫
∫
n n
ln
ln
ln
ln
csch
csch ln
csch ln
3
2
3
3
3
3
2
2
2
∫
= −
= −
( )−
( ))
(
= −
−
−
−
dx
x
e
e
e
l ln
ln
2
2
2
3 1
3
2
2 1
2
6
9 1
4
4 1
3
4
4
−






= −
−
−
−










= − −
− −
(
) = − −
−
e
3 3
7
12
( ) =
799.
sin
−
( )
1 1
4
Here’s the given problem:
cosh
sinh
ln
x
x
dx
9
2
0
2
−
∫
To find the antiderivative, begin by using the substitution u = sinh x so that du =
cosh x
 
dx. You can find the new limits of integration by noting that if x = ln 2, then
u
e
e
=
( )=
−
=
− =
−
sinh ln
ln
ln
2
2
2 1
2
2
3
4
2
2
, and if x = 0, then u
e e
=
( ) = − =
sinh 0
2
0
0
0
. With
these substitutions, you produce the following: