ICAI Notes: Differential and Integral Calculus- 5 | Mathematics for GRE Paper II PDF Download

 Page 1


TABLE OF INTEGRALS
15 Elementary integrals
All of these follow immediately from the table of derivatives. They should be memorized.
Z
cf(x) dx = c
Z
f(x) dx
Z
[f + g] dx = 
Z
f dx + 
Z
g dx
Z
c dx = cx + C
Z
dx = x + C
Z
x
r
dx =
x
r+1
r + 1
+ C (r6=1)
Z
1
x
dx = lnjxj + C
Z
e
x
dx = e
x
+ C
Z
ln x dx = x ln x x + C
The following are elementary integrals involving trigonometric functions.
Z
sin x dx = cos x + C
Z
cos x dx = sin x + C
Z
sec
2
x dx = tan x + C
Z
csc
2
x dx = cot x + C
Z
1
a
2
+ x
2
dx =
1
a
tan
1
x
a
+ C
Z
1
p
a
2
 x
2
dx = sin
1
x
a
+ C
Z
tanx dx = lnj cos xj + C
Z
cotx dx = lnj sin xj + C
Z
sec x dx = lnj tanx + sec xj + C
Z
csc x dx = lnj cotx csc xj + C
The following are elementary integrals involving hyperbolic functions (for details read the
Section on Hyperbolic Functions).
Z
sinh x dx = cosh x + C
Z
cosh x dx = sinh x + C
Z
sech
2
x dx = tanh x + C
Z
csch
2
x dx = coth x + C
Z
1
a
2
 x
2
dx =
1
a
tanh
1
x
a
+ C
Z
1
p
a
2
+ x
2
dx = sinh
1
x
a
+ C
Z
tanh x dx = ln(cosh x) + C
Z
coth x dx = lnj sinh xj + C
Z
sech x dx = tan
1
(sinh x) + C
Z
csch x dx = lnj cothx csch xj + C
Page 2


TABLE OF INTEGRALS
15 Elementary integrals
All of these follow immediately from the table of derivatives. They should be memorized.
Z
cf(x) dx = c
Z
f(x) dx
Z
[f + g] dx = 
Z
f dx + 
Z
g dx
Z
c dx = cx + C
Z
dx = x + C
Z
x
r
dx =
x
r+1
r + 1
+ C (r6=1)
Z
1
x
dx = lnjxj + C
Z
e
x
dx = e
x
+ C
Z
ln x dx = x ln x x + C
The following are elementary integrals involving trigonometric functions.
Z
sin x dx = cos x + C
Z
cos x dx = sin x + C
Z
sec
2
x dx = tan x + C
Z
csc
2
x dx = cot x + C
Z
1
a
2
+ x
2
dx =
1
a
tan
1
x
a
+ C
Z
1
p
a
2
 x
2
dx = sin
1
x
a
+ C
Z
tanx dx = lnj cos xj + C
Z
cotx dx = lnj sin xj + C
Z
sec x dx = lnj tanx + sec xj + C
Z
csc x dx = lnj cotx csc xj + C
The following are elementary integrals involving hyperbolic functions (for details read the
Section on Hyperbolic Functions).
Z
sinh x dx = cosh x + C
Z
cosh x dx = sinh x + C
Z
sech
2
x dx = tanh x + C
Z
csch
2
x dx = coth x + C
Z
1
a
2
 x
2
dx =
1
a
tanh
1
x
a
+ C
Z
1
p
a
2
+ x
2
dx = sinh
1
x
a
+ C
Z
tanh x dx = ln(cosh x) + C
Z
coth x dx = lnj sinh xj + C
Z
sech x dx = tan
1
(sinh x) + C
Z
csch x dx = lnj cothx csch xj + C
16 A selection of more complicated integrals
These begin with the two basic formulas, change of variables and integration by parts.
Z
f(g(x))g
0
(x) dx =
Z
f(u) du where u = g(x) (change of variables)
Z
f(g(x)) dx =
Z
f(u)
dx
du
du where u = g(x) (dierent form of the same change of variables)
Z
e
cx
dx =
1
c
e
cx
+ C (c6= 0)
Z
a
x
dx =
1
ln a
a
x
+ C (for a > 0; a6= 1)
Z
ln x dx = x lnx x + C
Z
1
x
2
+ a
2
dx =
1
a
arctan
x
a
+ C; a6= 0
Z
1
x
2
 a
2
dx =
1
2a
ln



x a
x + a


 +C; a6= 0
Z
1
p
a
2
 x
2
dx = arcsin
x
a
+ C; a > 0
Z
1
p
x
2
 a
2
dx = lnjx +
p
x
2
 a
2
j + C
To compute
Z
1
x
2
+ bx + c
dx we complete the square
x
2
+ bx + c = x
2
+ bx +
b
2
4
+ c
b
2
4
=

x +
b
2

2
+ c
b
2
4
If c b
2
=4 > 0, set it equal to a
2
; if < 0 equal toa
2
; and if = 0 forget it. In any event you
will arrive after the change of variables u = x +
b
2
at one of the three integrals
Z
1
u
2
+ a
2
du;
Z
1
u
2
 a
2
du;
Z
1
u
2
du
Z
p
x
2
 a
2
dx =
1
2

x
p
x
2
 a
2
 a
2
ln


 x +
p
x
2
 a
2




+ C
Z
x
n
e
cx
dx = x
n
e
cx
c

n
c
Z
x
n1
e
cx
dx etc. This is to be used repetaedly until you arrive at the
case n = 0, which you can do easily.
Page 3


TABLE OF INTEGRALS
15 Elementary integrals
All of these follow immediately from the table of derivatives. They should be memorized.
Z
cf(x) dx = c
Z
f(x) dx
Z
[f + g] dx = 
Z
f dx + 
Z
g dx
Z
c dx = cx + C
Z
dx = x + C
Z
x
r
dx =
x
r+1
r + 1
+ C (r6=1)
Z
1
x
dx = lnjxj + C
Z
e
x
dx = e
x
+ C
Z
ln x dx = x ln x x + C
The following are elementary integrals involving trigonometric functions.
Z
sin x dx = cos x + C
Z
cos x dx = sin x + C
Z
sec
2
x dx = tan x + C
Z
csc
2
x dx = cot x + C
Z
1
a
2
+ x
2
dx =
1
a
tan
1
x
a
+ C
Z
1
p
a
2
 x
2
dx = sin
1
x
a
+ C
Z
tanx dx = lnj cos xj + C
Z
cotx dx = lnj sin xj + C
Z
sec x dx = lnj tanx + sec xj + C
Z
csc x dx = lnj cotx csc xj + C
The following are elementary integrals involving hyperbolic functions (for details read the
Section on Hyperbolic Functions).
Z
sinh x dx = cosh x + C
Z
cosh x dx = sinh x + C
Z
sech
2
x dx = tanh x + C
Z
csch
2
x dx = coth x + C
Z
1
a
2
 x
2
dx =
1
a
tanh
1
x
a
+ C
Z
1
p
a
2
+ x
2
dx = sinh
1
x
a
+ C
Z
tanh x dx = ln(cosh x) + C
Z
coth x dx = lnj sinh xj + C
Z
sech x dx = tan
1
(sinh x) + C
Z
csch x dx = lnj cothx csch xj + C
16 A selection of more complicated integrals
These begin with the two basic formulas, change of variables and integration by parts.
Z
f(g(x))g
0
(x) dx =
Z
f(u) du where u = g(x) (change of variables)
Z
f(g(x)) dx =
Z
f(u)
dx
du
du where u = g(x) (dierent form of the same change of variables)
Z
e
cx
dx =
1
c
e
cx
+ C (c6= 0)
Z
a
x
dx =
1
ln a
a
x
+ C (for a > 0; a6= 1)
Z
ln x dx = x lnx x + C
Z
1
x
2
+ a
2
dx =
1
a
arctan
x
a
+ C; a6= 0
Z
1
x
2
 a
2
dx =
1
2a
ln



x a
x + a


 +C; a6= 0
Z
1
p
a
2
 x
2
dx = arcsin
x
a
+ C; a > 0
Z
1
p
x
2
 a
2
dx = lnjx +
p
x
2
 a
2
j + C
To compute
Z
1
x
2
+ bx + c
dx we complete the square
x
2
+ bx + c = x
2
+ bx +
b
2
4
+ c
b
2
4
=

x +
b
2

2
+ c
b
2
4
If c b
2
=4 > 0, set it equal to a
2
; if < 0 equal toa
2
; and if = 0 forget it. In any event you
will arrive after the change of variables u = x +
b
2
at one of the three integrals
Z
1
u
2
+ a
2
du;
Z
1
u
2
 a
2
du;
Z
1
u
2
du
Z
p
x
2
 a
2
dx =
1
2

x
p
x
2
 a
2
 a
2
ln


 x +
p
x
2
 a
2




+ C
Z
x
n
e
cx
dx = x
n
e
cx
c

n
c
Z
x
n1
e
cx
dx etc. This is to be used repetaedly until you arrive at the
case n = 0, which you can do easily.
17 Hyperbolic Functions
Denition: The most frequently used hyperbolic functions are dened by,
sinh x =
e
x
 e
x
2
cosh x =
e
x
+ e
x
2
tanh x =
sinh x
cosh x
=
e
x
 e
x
e
x
+ e
x
where sinh x and tanh x are odd like sin x and tanx, while cosh x is even like cos x (see Fig. 4
for plots of these functions). The denitions of other hyperbolic functions are identical to the
corresponding trig functions.
-4 0 4
-10
10
sinh(x)
-3 -2 -1 0 1 2 3
3
6
cosh(x)
-4 -2 0 2 4
-1
1
tanh(x)
Figure 4: Graphs of sinh x, cosh x, and tanhx.
We can easily derive all the corresponding identities for hyperbolic functions by using the
denition. However, it helps us memorize these identities if we know the relationship between
sinh x, cosh x and sin x, cos x in complex analysis.
Combining sinh x =
e
x
 e
x
2
and sin x =
e
ix
 e
ix
2i
, we obtain sinh(x) =i sin(ix):
Combining cosh x =
e
x
+ e
x
2
and cos x =
e
ix
+ e
ix
2
, we obtain cosh(x) = cos(ix):
where the constant i =
p
1 and i
2
=1. Now, we can write down these identities:
cos
2
x + sin
2
x = 1 cosh
2
x sinh
2
x = 1
1 + tan
2
x = sec
2
x 1 tanh
2
x = sech
2
x
cot
2
x + 1 = csc
2
x coth
2
x 1 = csch
2
x
cos
2
x = (1 + cos 2x)=2 cosh
2
x = (1 + cosh 2x)=2
sin
2
x = 1 cos
2
x = (1 cos 2x)=2 sinh
2
x = cosh
2
x 1 = (1 + cosh 2x)=2
sin 2x = 2 sin x cos x sinh 2x = 2 sinh x cosh x
sin(x y) = sin x cos y cos x sin y sinh(x y) = sinh x cosh y cosh x sinh y
cos(x y) = cos x cos y sin x sin y cosh(x y) = cosh x cosh y sinh x sinh y
Page 4


TABLE OF INTEGRALS
15 Elementary integrals
All of these follow immediately from the table of derivatives. They should be memorized.
Z
cf(x) dx = c
Z
f(x) dx
Z
[f + g] dx = 
Z
f dx + 
Z
g dx
Z
c dx = cx + C
Z
dx = x + C
Z
x
r
dx =
x
r+1
r + 1
+ C (r6=1)
Z
1
x
dx = lnjxj + C
Z
e
x
dx = e
x
+ C
Z
ln x dx = x ln x x + C
The following are elementary integrals involving trigonometric functions.
Z
sin x dx = cos x + C
Z
cos x dx = sin x + C
Z
sec
2
x dx = tan x + C
Z
csc
2
x dx = cot x + C
Z
1
a
2
+ x
2
dx =
1
a
tan
1
x
a
+ C
Z
1
p
a
2
 x
2
dx = sin
1
x
a
+ C
Z
tanx dx = lnj cos xj + C
Z
cotx dx = lnj sin xj + C
Z
sec x dx = lnj tanx + sec xj + C
Z
csc x dx = lnj cotx csc xj + C
The following are elementary integrals involving hyperbolic functions (for details read the
Section on Hyperbolic Functions).
Z
sinh x dx = cosh x + C
Z
cosh x dx = sinh x + C
Z
sech
2
x dx = tanh x + C
Z
csch
2
x dx = coth x + C
Z
1
a
2
 x
2
dx =
1
a
tanh
1
x
a
+ C
Z
1
p
a
2
+ x
2
dx = sinh
1
x
a
+ C
Z
tanh x dx = ln(cosh x) + C
Z
coth x dx = lnj sinh xj + C
Z
sech x dx = tan
1
(sinh x) + C
Z
csch x dx = lnj cothx csch xj + C
16 A selection of more complicated integrals
These begin with the two basic formulas, change of variables and integration by parts.
Z
f(g(x))g
0
(x) dx =
Z
f(u) du where u = g(x) (change of variables)
Z
f(g(x)) dx =
Z
f(u)
dx
du
du where u = g(x) (dierent form of the same change of variables)
Z
e
cx
dx =
1
c
e
cx
+ C (c6= 0)
Z
a
x
dx =
1
ln a
a
x
+ C (for a > 0; a6= 1)
Z
ln x dx = x lnx x + C
Z
1
x
2
+ a
2
dx =
1
a
arctan
x
a
+ C; a6= 0
Z
1
x
2
 a
2
dx =
1
2a
ln



x a
x + a


 +C; a6= 0
Z
1
p
a
2
 x
2
dx = arcsin
x
a
+ C; a > 0
Z
1
p
x
2
 a
2
dx = lnjx +
p
x
2
 a
2
j + C
To compute
Z
1
x
2
+ bx + c
dx we complete the square
x
2
+ bx + c = x
2
+ bx +
b
2
4
+ c
b
2
4
=

x +
b
2

2
+ c
b
2
4
If c b
2
=4 > 0, set it equal to a
2
; if < 0 equal toa
2
; and if = 0 forget it. In any event you
will arrive after the change of variables u = x +
b
2
at one of the three integrals
Z
1
u
2
+ a
2
du;
Z
1
u
2
 a
2
du;
Z
1
u
2
du
Z
p
x
2
 a
2
dx =
1
2

x
p
x
2
 a
2
 a
2
ln


 x +
p
x
2
 a
2




+ C
Z
x
n
e
cx
dx = x
n
e
cx
c

n
c
Z
x
n1
e
cx
dx etc. This is to be used repetaedly until you arrive at the
case n = 0, which you can do easily.
17 Hyperbolic Functions
Denition: The most frequently used hyperbolic functions are dened by,
sinh x =
e
x
 e
x
2
cosh x =
e
x
+ e
x
2
tanh x =
sinh x
cosh x
=
e
x
 e
x
e
x
+ e
x
where sinh x and tanh x are odd like sin x and tanx, while cosh x is even like cos x (see Fig. 4
for plots of these functions). The denitions of other hyperbolic functions are identical to the
corresponding trig functions.
-4 0 4
-10
10
sinh(x)
-3 -2 -1 0 1 2 3
3
6
cosh(x)
-4 -2 0 2 4
-1
1
tanh(x)
Figure 4: Graphs of sinh x, cosh x, and tanhx.
We can easily derive all the corresponding identities for hyperbolic functions by using the
denition. However, it helps us memorize these identities if we know the relationship between
sinh x, cosh x and sin x, cos x in complex analysis.
Combining sinh x =
e
x
 e
x
2
and sin x =
e
ix
 e
ix
2i
, we obtain sinh(x) =i sin(ix):
Combining cosh x =
e
x
+ e
x
2
and cos x =
e
ix
+ e
ix
2
, we obtain cosh(x) = cos(ix):
where the constant i =
p
1 and i
2
=1. Now, we can write down these identities:
cos
2
x + sin
2
x = 1 cosh
2
x sinh
2
x = 1
1 + tan
2
x = sec
2
x 1 tanh
2
x = sech
2
x
cot
2
x + 1 = csc
2
x coth
2
x 1 = csch
2
x
cos
2
x = (1 + cos 2x)=2 cosh
2
x = (1 + cosh 2x)=2
sin
2
x = 1 cos
2
x = (1 cos 2x)=2 sinh
2
x = cosh
2
x 1 = (1 + cosh 2x)=2
sin 2x = 2 sin x cos x sinh 2x = 2 sinh x cosh x
sin(x y) = sin x cos y cos x sin y sinh(x y) = sinh x cosh y cosh x sinh y
cos(x y) = cos x cos y sin x sin y cosh(x y) = cosh x cosh y sinh x sinh y
The Advantage of Hyperbolic Functions
The advantage of hyperbolic functions is that their inverse functions can be expressed in terms
of elementary functions. We here show how to solve for the inverse hyperbolic functions.
By denition, u = sinh x =
e
x
 e
x
2
.
Multiplying both sides by 2e
x
, we obtain
2ue
x
= (e
x
)
2
 1 which simplies to (e
x
)
2
 2u(e
x
) 1 = 0.
This is a quadratic in e
x
, Thus,
e
x
= [2u
p
(2u)
2
+ 4]=2 = u
p
u
2
+ 1.
The negative sign in does not make sense since e
x
> 0 for all x.
Thus, x = ln e
x
= lnju +
p
u
2
+ 1j which implies
x = sinh
1
u = lnju +
p
u
2
+ 1j:
Similarly, by denition, u = cosh x =
e
x
+ e
x
2
.
We can solve this equation for the two branches of the inverse:
x = cosh
1
u = lnju
p
u
2
 1j = lnju +
p
u
2
 1j.
Similarly, for
u = tanhx =
e
x
 e
x
e
x
+ e
x
.
We can solve the equation for the inverse:
x = tanh
1
u =
1
2
lnj
1 + u
1 u
j.
Page 5


TABLE OF INTEGRALS
15 Elementary integrals
All of these follow immediately from the table of derivatives. They should be memorized.
Z
cf(x) dx = c
Z
f(x) dx
Z
[f + g] dx = 
Z
f dx + 
Z
g dx
Z
c dx = cx + C
Z
dx = x + C
Z
x
r
dx =
x
r+1
r + 1
+ C (r6=1)
Z
1
x
dx = lnjxj + C
Z
e
x
dx = e
x
+ C
Z
ln x dx = x ln x x + C
The following are elementary integrals involving trigonometric functions.
Z
sin x dx = cos x + C
Z
cos x dx = sin x + C
Z
sec
2
x dx = tan x + C
Z
csc
2
x dx = cot x + C
Z
1
a
2
+ x
2
dx =
1
a
tan
1
x
a
+ C
Z
1
p
a
2
 x
2
dx = sin
1
x
a
+ C
Z
tanx dx = lnj cos xj + C
Z
cotx dx = lnj sin xj + C
Z
sec x dx = lnj tanx + sec xj + C
Z
csc x dx = lnj cotx csc xj + C
The following are elementary integrals involving hyperbolic functions (for details read the
Section on Hyperbolic Functions).
Z
sinh x dx = cosh x + C
Z
cosh x dx = sinh x + C
Z
sech
2
x dx = tanh x + C
Z
csch
2
x dx = coth x + C
Z
1
a
2
 x
2
dx =
1
a
tanh
1
x
a
+ C
Z
1
p
a
2
+ x
2
dx = sinh
1
x
a
+ C
Z
tanh x dx = ln(cosh x) + C
Z
coth x dx = lnj sinh xj + C
Z
sech x dx = tan
1
(sinh x) + C
Z
csch x dx = lnj cothx csch xj + C
16 A selection of more complicated integrals
These begin with the two basic formulas, change of variables and integration by parts.
Z
f(g(x))g
0
(x) dx =
Z
f(u) du where u = g(x) (change of variables)
Z
f(g(x)) dx =
Z
f(u)
dx
du
du where u = g(x) (dierent form of the same change of variables)
Z
e
cx
dx =
1
c
e
cx
+ C (c6= 0)
Z
a
x
dx =
1
ln a
a
x
+ C (for a > 0; a6= 1)
Z
ln x dx = x lnx x + C
Z
1
x
2
+ a
2
dx =
1
a
arctan
x
a
+ C; a6= 0
Z
1
x
2
 a
2
dx =
1
2a
ln



x a
x + a


 +C; a6= 0
Z
1
p
a
2
 x
2
dx = arcsin
x
a
+ C; a > 0
Z
1
p
x
2
 a
2
dx = lnjx +
p
x
2
 a
2
j + C
To compute
Z
1
x
2
+ bx + c
dx we complete the square
x
2
+ bx + c = x
2
+ bx +
b
2
4
+ c
b
2
4
=

x +
b
2

2
+ c
b
2
4
If c b
2
=4 > 0, set it equal to a
2
; if < 0 equal toa
2
; and if = 0 forget it. In any event you
will arrive after the change of variables u = x +
b
2
at one of the three integrals
Z
1
u
2
+ a
2
du;
Z
1
u
2
 a
2
du;
Z
1
u
2
du
Z
p
x
2
 a
2
dx =
1
2

x
p
x
2
 a
2
 a
2
ln


 x +
p
x
2
 a
2




+ C
Z
x
n
e
cx
dx = x
n
e
cx
c

n
c
Z
x
n1
e
cx
dx etc. This is to be used repetaedly until you arrive at the
case n = 0, which you can do easily.
17 Hyperbolic Functions
Denition: The most frequently used hyperbolic functions are dened by,
sinh x =
e
x
 e
x
2
cosh x =
e
x
+ e
x
2
tanh x =
sinh x
cosh x
=
e
x
 e
x
e
x
+ e
x
where sinh x and tanh x are odd like sin x and tanx, while cosh x is even like cos x (see Fig. 4
for plots of these functions). The denitions of other hyperbolic functions are identical to the
corresponding trig functions.
-4 0 4
-10
10
sinh(x)
-3 -2 -1 0 1 2 3
3
6
cosh(x)
-4 -2 0 2 4
-1
1
tanh(x)
Figure 4: Graphs of sinh x, cosh x, and tanhx.
We can easily derive all the corresponding identities for hyperbolic functions by using the
denition. However, it helps us memorize these identities if we know the relationship between
sinh x, cosh x and sin x, cos x in complex analysis.
Combining sinh x =
e
x
 e
x
2
and sin x =
e
ix
 e
ix
2i
, we obtain sinh(x) =i sin(ix):
Combining cosh x =
e
x
+ e
x
2
and cos x =
e
ix
+ e
ix
2
, we obtain cosh(x) = cos(ix):
where the constant i =
p
1 and i
2
=1. Now, we can write down these identities:
cos
2
x + sin
2
x = 1 cosh
2
x sinh
2
x = 1
1 + tan
2
x = sec
2
x 1 tanh
2
x = sech
2
x
cot
2
x + 1 = csc
2
x coth
2
x 1 = csch
2
x
cos
2
x = (1 + cos 2x)=2 cosh
2
x = (1 + cosh 2x)=2
sin
2
x = 1 cos
2
x = (1 cos 2x)=2 sinh
2
x = cosh
2
x 1 = (1 + cosh 2x)=2
sin 2x = 2 sin x cos x sinh 2x = 2 sinh x cosh x
sin(x y) = sin x cos y cos x sin y sinh(x y) = sinh x cosh y cosh x sinh y
cos(x y) = cos x cos y sin x sin y cosh(x y) = cosh x cosh y sinh x sinh y
The Advantage of Hyperbolic Functions
The advantage of hyperbolic functions is that their inverse functions can be expressed in terms
of elementary functions. We here show how to solve for the inverse hyperbolic functions.
By denition, u = sinh x =
e
x
 e
x
2
.
Multiplying both sides by 2e
x
, we obtain
2ue
x
= (e
x
)
2
 1 which simplies to (e
x
)
2
 2u(e
x
) 1 = 0.
This is a quadratic in e
x
, Thus,
e
x
= [2u
p
(2u)
2
+ 4]=2 = u
p
u
2
+ 1.
The negative sign in does not make sense since e
x
> 0 for all x.
Thus, x = ln e
x
= lnju +
p
u
2
+ 1j which implies
x = sinh
1
u = lnju +
p
u
2
+ 1j:
Similarly, by denition, u = cosh x =
e
x
+ e
x
2
.
We can solve this equation for the two branches of the inverse:
x = cosh
1
u = lnju
p
u
2
 1j = lnju +
p
u
2
 1j.
Similarly, for
u = tanhx =
e
x
 e
x
e
x
+ e
x
.
We can solve the equation for the inverse:
x = tanh
1
u =
1
2
lnj
1 + u
1 u
j.
The Advantage of Trigonometric Functions
Relations between dierent trigonometric functions are very important since when we dif-
ferentiate or integrate one trig function we obtain another trig function. When we use trig
substitution, it is often a necessity for us to know the denition of other trig functions that is
related to the one we use in the substitution.
Example: Calculate the derivative of y = sin
1
x.
Solution: y = sin
1
x means sin y = x. Dierentiate both sides of sin y = x, we obtain
cos yy
0
= 1 ) y
0
=
1
cos y
:
In order to express y
0
in terms of x, we need to express cos y in terms of x. Given that sin y = x,
we can obtain cos y =
p
cos
2
y =
p
1 sin
2
y =
p
1 x
2
by using trig identities. However, it
is easier to construct a right triangle with an angle y. The opposite side must be x while the
hypotenuse must be 1, thus the adjacent side is
p
1 x
2
. Therefore,
cos y =
adjacent side
hypotenuse
=
p
1 x
2
1
=
p
1 x
2
:
Example: Calculate the integral
R p
1 + x
2
dx.
Solution: This integral requires standard trig substitution x = tanu or x = sinh u. Let's use
x = sinh u. Recall that 1 + sinh
2
u = cosh
2
u and that dx = d sinh u = cosh udu,
Z
p
1 + x
2
dx =
Z
p
1 + sinh
2
ud sinh u =
Z
cosh
2
udu
=
1
2
Z
[1 + cosh(2u)]du =
1
2
[u +
1
2
sinh(2u)] + C =
1
2
[u + sinh u cosh u] + C:
where hyperbolic identities cosh
2
u = [1 + cosh(2u)]=2 and sinh(2u) = 2 sinh u coshu were
used. However, we need to express the solution in terms of x. Since x = sinh u was the
substitution, we know right away u = sinh
1
x = lnjx +
p
1 + x
2
j and sinh u = x, but how to
express cosh u in terms of x? We can solve it using hyperbolic identities. cosh u =
p
cosh
2
u =
p
1 + sinh
2
u =
p
1 + x
2
. Thus,
1
2
[u + sinh u coshu] =
1
2
[lnjx +
p
1 + x
2
j + x
p
1 + x
2
]: