Integral Calculus 201-NYB-05
Vincent Carrier
Basic Formulas
Let f be a real-valued function and F an antiderivative of f . The problem of finding F
is not simple in general. A special case where F can be found without too much effort is
the case where f is the derivative of a well-known function.
Examples: Find an antiderivative F .
a) f (x) = cos x
F (x) = sin x
b) f (x) = ex
c) f (x) = − csc2 x
F (x) = cot x
d) f (x) =
1
x
F (x) = ln |x|
1
e) f (x) = √
1 − x2
F (x) = arcsin x
f)
1
1 + x2
F (x) = arctan x
f (x) =
F (x) = ex
Consider the case where f (x) = xn for n ∈ R. The Power Rule states that
f 0 (x) = nxn−1 .
If we apply the Power Rule in the reversed direction, we get
F (x) =
xn+1
n+1
for n 6= −1.
Examples: Find an antiderivative F .
F (x) =
x4
4
F (x) =
x−7
1
= − 7
−7
7x
1
f (x) = √
= x−7/3
3
7
x
F (x) =
x−4/3
3
= − 4/3
−4/3
4x
√
3
x x2
x5/3
f (x) = √
= 3/2 = x1/6
x
x3
F (x) =
x7/6
6x7/6
=
7/6
7
a)
f (x) = x3
b)
f (x) =
c)
d)
1
= x−8
8
x
Sometimes, the above technique can be applied even if f is not exactly the derivative of
a well-known function.
Assume that we are looking for an antiderivative of c f (ax + b) where a, b, c ∈ R and that
F (x) is an antiderivative of f (x). Then
c F (ax + b)
a
is an antiderivative of
c f (ax + b).
Indeed,
d c F (ax + b)
c
= [F 0 (ax + b)(a)] = c f (ax + b).
dx
a
a
Examples: Find an antiderivative F .
a)
f (x) = e4x+3
b)
f (x) =
c)
f (x) = 3 sec2
d)
f (x) = √
6
= 6(4 − 3x)−2
(4 − 3x)2
x
2
5
5
= p
1 − 4x2
1 − (2x)2
F (x) =
e4x+3
4
F (x) =
6(4 − 3x)−1
2
=
(−1)(−3)
4 − 3x
F (x) =
x
3 tan(x/2)
= 6 tan
1/2
2
F (x) =
5 arcsin 2x
2
Some examples require a more sophisticated application of the Chain Rule
[f (g(x))]0 = f 0 (g(x)) g 0 (x).
Examples: Find an antiderivative F .
3
a)
f (x) = 3x2 ex
b)
f (x) = cot x =
c)
f (x) =
3
F (x) = ex
cos x
sin x
2x
2x
=
4
1+x
1 + (x2 )2
F (x) = ln | sin x|
F (x) = arctan x2