Things to remember from Calculus 1.
Derivatives:
f ′ (x)
f (x)
C
0
xp
p xp−1
1
x
ex
cos x
log x
ex
sin x
cos x
− sin x
sec2 x
tan x
sec x
cosec x
sec x tan x
− cosec x cot x
− cosec2 x
1
√
1 − x2
1
−√
1 − x2
1
1 + x2
cot x
arcsin x
arccos x
arctan x
Product Rule:
f (x)g(x)
f ′ (x)g(x) + f (x)g ′ (x)
Quotient Rule:
f (x)
g(x)
f ′ (x)g(x) − f (x)g ′ (x)
[g(x)]2
Chain Rule:
f ′ (g(x))g ′(x)
f (g(x))
Exercise: Derive all the trigonometric relations from just one. The derivative of sin x is
cos x.
π
π
π
(cos x)′ = [sin( − x)]′ = [cos( − x)] × ( − x)′ = [sin x] × (−1) = − sin x
2
2
2
′
′
(sin x) (cos x) − (sin x)(cos x)′
sin x
′
=
(tan x) =
cos x
cos2 x
1
(cos x)2 + (sin x)2
=
=
2
cos x
cos2 x
= sec2 x
1
Anti-derivatives, Indefinite Integrals.
f ′ (x)
Z
Z
Z
Z
f (x) + C
f ′ (x)dx = f (x) + C
Z
xk dx
xk+1
+C
k+1
Z
1
dx
x
log x + C
Z
cos x dx
sin x + C
Z
sin x dx
− cos x + C
Z
tan x dx
− log cos x + C
1
√
dx
1 − x2
Z
1
dx
1 + x2
Z
ex dx
Z ′
f (x)
dx
f (x)
if k 6= 1
arcsin x + C
arctan x + C
ex + C
log f (x) + C
f (x)k+1
+C
k+1
f (x)k f ′ (x) dx
f ′ (g(x))g ′(x) dx
f (g(x)) + C
Substitution.
Z
Z
Z
dx
f (x)dx = F (x) + C; x = g(y); f (g(y)) dy = f (g(y))g ′(y)dy = F (g(y)) + C
dy
Examples:
Z
√
1
dx
1 − x2
Put x = sin y.
Z
Z
Z
1
cos y
√
p
dx =
dy = 1 dy = y + C = arcsin x + C
1 − x2
1 − sin2 y
Z
1
dx
1 + x2
2
Put x = tan y.
Z
Z
Z
1
sec2 y
dy = 1 dy = y + C = arctan x + C
dx =
1 + x2
1 + tan2 y
R √
R √
In general for f ( 1 − x2 )dx try x = sin y and for f ( 1 + x2 ) try x = tan y.
Integration by parts.
Z
or
′
u(x)v (x)dx = u(x)v(x) −
Z
u(x) dv(x) = u(x)v(x) −
Z
Z
Z
v(x)u′ (x)dx
Z
v(x) du(x)
Examples:
Z
Z
log x dx =
x sin x dx = −
Z
log x · 1 dx = x log x −
x d cos x = −x cos x +
Z
x·
1
dx = xlogx − x + C
x
cos x dx = −x cos x + sin x + C
You can integrate by parts repeatedly.
Z
Z
Z
Z
2 x
2 x
2 x
x
2 x
x
x e dx = x de = x e − 2 xe dx = x e − 2xe + 2 ex dx = (x2 − 2x + 2)ex + C
Definite Integrals. Fundamental Theorem of Calculus. Area under a curve.
If y = f (x) is a curve (assume f (x) > 0) then the area A between the curve and the x-axis
between the ordinates x = a and x = b can be obtained as
Z b
f (x)dx = F (b) − F (a)
A=
a
where F ′ (x) = f (x). In particular if
A(y) =
Z
y
f (x)dx
a
then
A′ (y) = f (y)
If f is not positive then the area above x-axis is positive and the area below is negative.
Rb
Then the sum is what can still be computed as a f (x)dx. For example,
0
x2 1
x dx =
=−
2 −1
2
−1
Z
0
3
Z
1
0
and
1
1
x2 =
x dx =
2 0 2
1
x2 x dx =
=0
2 −1
−1
Z
1
By convention to be consistent we define for a > b,
Z
b
f (x)dx = −
a
Z
a
f (x)dx
b
so that for any a, b
Z
b
a
f (x)dx = F (b) − F (a)
Examples: Area inside a circle of radius r.
A(r) = 4
Z
r
0
p
r 2 − x2 dx
Put x = r sin y. dx = r cos ydy. Change limits. when x = 0, y = 0. When x = r, y =
A(r) = 4
Z
0
π
2
r cos y · r cos y dy = 4 r
2
Z
π
2
2
cos ydy = 2r
0
π
sin 2y 2
] = πr 2
= 2r [y −
2 0
2
Why should you change limits? Chain rule says
[f (g(x))]′ = f ′ (g(x))g ′(x)
In particular
Z
b
a
[f (g(x))]′dx = f (g(b)) − f (g(a))
If we change variables by y = g(x) the integral becomes
Z
f ′ (y)dy = f (y) + C
And
Z
g(b)
g(a)
f ′ (y)dy = f (g(b)) − f (g(a))
is the right answer.
4
2
Z
0
π
2
(1 − cos 2y)dy
π
2.
Partial Fractions. This involves rewriting the function f to be integrated in a different
form to make it easier.
Examples:
Z
Write
1
dx
(x + 1)(x + 2)
1
1
1
=
−
(x + 1)(x + 2)
x+1 x+2
Then
Z
Z
Z
1
1
1
dx =
dx −
dx
(x + 1)(x + 2)
x+1
x+2
= log(x + 1) − log(x + 2) + C
Another Example.
Z
3x + 4
dx
(x + 1)(x + 2)
a
b
(a + b)x + (2a + b)
3x + 4
=
+
=
(x + 1)(x + 2)
x+1 x+2
(x + 1)(x + 2)
a + b = 3, 2a + b = 4
a = 1, b = 2
1
2
3x + 4
=
+
(x + 1)(x + 2)
x+1 x+2
Z
3x + 4
dx = log(x + 1) + 2 log(x + 2) + C
(x + 1)(x + 2)
5