Part II: The Answers
486
Answers
701–800
To find the antiderivative, use the substitution u = 1 – 3x so that du = –3
 
dx,
or −
=
1
3
du dx:
sinh
sinh
cosh
cosh
1 3
1
3
1
3
1
3
1 3
−
(
) = −
( )
= −
+
= −
−
(
)+
∫
∫
x dx
udu
u C
x C
791.
1
3
3
3
cosh x
C
−
(
)+
Here’s the given problem:
cosh
sinh
2
3
3
x
x
dx
−
(
)
−
(
)
∫
To find the antiderivative, begin by using the substitution u = cosh (x – 3) so that
du = sinh (x – 3)
 
dx. With this substitution, you get the following:
cosh
sinh
cosh
2
2
3
3
3
3
3
1
3
3
x
x
dx
u du
u C
x
C
−
(
)
−
(
)
=
=
+
=
−
(
)+
∫
∫
792.
ln sinh x C
+
Rewrite the problem using the definition of hyperbolic cotangent:
coth
cosh
sinh
x dx
x
x
dx
∫
∫
=
Begin finding the antiderivative by using the substitution u = sinh x so that
du = cosh x
 
dx. With these substitutions, you get the following:
1
u
du
u C
x C
=
+ =
+
∫
ln
ln sinh
793.
1
3
3
2
tanh x
C
−
(
)+
Here’s the given problem:
sech
2 3
2
x
dx
−
(
)
∫
To find the antiderivative, use the substitution u = 3x – 2 to get du = 3
 
dx so that
1
3
du dx
= . With these substitutions, you get the following:
1
3
1
3
1
3
3
2
2
sech
tanh
tanh
u du
u C
x
C
( ) =
( )+
=
−
(
)+
∫