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# Common Derivatives Integrals

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```Common Derivatives and Integrals
Derivatives
Basic Properties/Formulas/Rules
d
cf (x) = cf 0 (x), c is any constant.
dx
d n
x = nxn−1 , n is any number.
dx
0
f (x) g(x) = f 0 (x) g(x) + f (x) g 0 (x) – Product Rule
f (x) 0 f 0 (x) g(x) − f (x) g 0 (x)
=
– Quotient Rule
2
g(x)
g(x)
d
f g(x) = f 0 g(x) g 0 (x) – Chain Rule
dx
d
f (x) &plusmn; g(x) = f 0 (x) &plusmn; g 0 (x)
dx
d c = 0, c is any constant.
dx
d g(x) e
= g 0 (x)eg(x)
dx
d
g 0 (x)
ln(g(x)) =
dx
g(x)
Common Derivatives
Polynomials
d c =0
dx
d
(x) = 1
dx
d cx = c
dx
Trig Functions
i
dh
sin(x) = cos(x)
dx
i
dh
csc(x) = − csc(x) cot(x)
dx
Inverse Trig Functions
d h −1 i
1
sin (x) = √
dx
1 − x2
h
i
d
1
csc−1 (x) = − √
dx
|x| x2 − 1
Exponential &amp; Logarithm Functions
d h xi
a = ax ln(a)
dx
i 1
dh
ln(x) = , x &gt; 0
dx
x
d n
x = nxn−1
dx
i
dh
cos(x) = − sin(x)
dx
i
dh
sec(x) = sec(x) tan(x)
dx
i
dh
1
cos−1 (x) = − √
dx
1 − x2
h
i
d
1
√
sec−1 (x) =
dx
|x| 1 − x2
d h xi
e = ex
dx
i 1
dh
ln |x| = , x 6= 0
dx
x
Hyperbolic Functions
i
dh
sinh(x) = cosh(x)
dx
i
dh
csch(x) = − csch(x) coth(x)
dx
d n
cx = ncxn−1
dx
i
dh
tan(x) = sec2 (x)
dx
i
dh
cot(x) = − csc2 (x)
dx
i
dh
1
tan−1 (x) =
dx
1 + x2
i
dh
1
cot−1 (x) = −
dx
1 + x2
i
dh
1
loga (x) =
, x&gt;0
dx
x ln(a)
i
dh
cosh(x) = sinh(x)
dx
i
dh
sech(x) = − sech(x) tanh(x)
dx
i
dh
tanh(x) = sech2 (x)
dx
i
dh
coth(x) = − csch2 (x)
dx
&copy; Paul Dawkins - https://tutorial.math.lamar.edu
Common Derivatives and Integrals
Integrals
Basic Properties/Formulas/Rules
Z
Z
cf (x) dx = c f (x) dx, c is a constant.
Z
a
b
f (x) dx
b
Z
Z
f (x) &plusmn; g(x) dx =
a
a
a
b
Z
Z
f (x) dx = −
a
a
b
Z
f (x) dx =
c
Z
b
a
Z
f (x) dx &plusmn;
b
g(x) dx
a
a
f (x) dx
b
b
Z
f (x) dx +
b
a
f (x) dx = 0
a
g(x) dx
Z
f (x) dx, c is a constant.
cf (x) dx = c
Z
f (x)dx &plusmn;
a
b
Z
Z
f (x) &plusmn; g(x) dx =
= F (b) − F (a) where f (x) =
f (x) dx = f (x)
a
Z
Z
b
b
Z
Z
c dx = c(b − a), c is a constant.
f (x) dx
c
a
Z
b
If f (x) ≥ 0 on a ≤ x ≤ b then
f (x) dx ≥ 0
a
b
Z
If f (x) ≥ g(x) on a ≤ x ≤ b then
Z
f (x) dx ≥
a
b
g(x) dx
a
Common Integrals
Polynomials
Z
Z
Z
1
dx = x + c
k dx = kx + c
xn dx =
xn+1 + c, n 6= −1
n+1
Z
Z
Z
1
1
−1
dx = ln |x| + c
x dx = ln |x| + c
x−n dx =
x−n+1 + c, n 6= 1
x
−n + 1
Z
Z
p
p
p+q
1
1
1
q
+1
dx = ln |ax + b| + c
x q dx = p
xq + c =
x q +c
ax + b
a
p+q
q +1
Trig Functions
Z
cos(u) du = sin(u) + c
Z
sec(u) tan(u) du = sec(u) + c
Z
Z
sec2 u du = tan(u) + c
Z
csc2 u du = − cot(u) + c
sin(u) du = − cos(u) + c
Z
csc(u) cot(u) du = − csc(u) + c
Z
Z
tan(u) du = − ln cos(u) +c = ln sec(u) +c
Z
1
sec (u) du =
sec(u) tan(u)+ln sec(u)+tan(u) +c
2
Z
1
3
csc (u) du =
−csc(u) cot(u)+ln csc(u)−cot(u) +c
2
Z
sec(u) du = ln sec(u)+tan(u) +c
Z
csc(u) du = ln csc(u)−cot(u) +c
cot(u) du = ln sin(u) +c = − ln csc(u) +c
3
&copy; Paul Dawkins - https://tutorial.math.lamar.edu
Common Derivatives and Integrals
Exponential &amp; Logarithm Functions
Z
Z
au
u
u
e du = e + c
au du =
+c
ln(a)
Z
eau
au
e sin(bu) du = 2
a sin(bu) − b cos(bu) + c
a + b2
Z
eau
au
e cos(bu) du = 2
a cos(bu) + b sin(bu) + c
a + b2
Inverse Trig Functions
Z
u
1
√
du = sin−1
+c
a
a2 − u2
Z
1
1
−1 u
du
=
tan
+c
a2 + u2
a
a
Z
u
1
1
√
du = sec−1
+c
a
a
u u2 − a2
Hyperbolic Functions
Z
sinh(u)du = cosh(u) + c
Z
ln(u) du = u ln(u) − u + c
Z
ueu du = (u − 1)eu + c
Z
1
du = ln ln(u) + c
u ln(u)
Z
sin−1 (u) du = u sin−1 (u) +
Z
tan−1 (u) du = u tan−1 (u) −
Z
csch(u) coth(u)du = − csch(u) + c
Z
a
sech2 (u)du = tanh(u) + c
Z
csch2 (u)du = − coth(u) + c
sech(u)du = tan−1 sinh(u) + c
Miscellaneous
Z
Z p
1
1
u+a
up 2
2 + u2 du =
du
=
ln
+
c
a
a + u2 +
a2 − u2
2a
u−a
2
Z
Z p
1
1
u−a
up 2
2 − a2 du =
du
=
ln
+
c
u
u − a2 −
u2 − a2
2a
u+a
2
Z p
u
up 2
a2
a2 − u2 du =
a − u2 +
sin−1
+c
2
2
a
Z p
u − ap
a2
a−u
2au − u2 du =
2au − u2 +
cos−1
+c
2
2
a
u Substitution :
Z
sech(u) tanh(u)du = − sech(u) + c
tanh(u)du = ln cosh(u) + c
Z
p
1 − u2 + c
1
ln 1 + u2 + c
2
Z
p
cos−1 (u) du = u cos−1 (u) − 1 − u2 + c
Z
cosh(u)du = sinh(u) + c
Z
Z
p
a2
ln u + a2 + u2 + c
2
p
a2
ln u + u2 − a2 + c
2
Standard Integration Techniques
Z b
Z
b
0
0
f g(x) g (x) dx will convert the integral into
f g(x) g (x) dx =
a
g(b)
g(a)
f (u) du
using the substitution u = g(x) where du = g 0 (x)dx. For indefinite integrals drop the limits of
integration.
&copy; Paul Dawkins - https://tutorial.math.lamar.edu
Common Derivatives and Integrals
Z
Integration by Parts :
Z
u dv = uv −
Z
b
b
v du and
Z
b
−
v du. Choose u and dv from
Z
integral and compute du by differentiating u and compute v using v = dv.
u dv = uv
a
a
a
Trig Substitutions : If the integral contains the following root use the given substitution and formula.
p
a2 − b2 x2
p
b2 x2 − a2
p
a2 + b2 x2
⇒
⇒
⇒
a
sin(θ)
b
a
x = sec(θ)
b
a
x = tan(θ)
b
x=
and
cos2 (θ) = 1 − sin2 (θ)
and
tan2 (θ) = sec2 (θ) − 1
and
sec2 (θ) = 1 + tan2 (θ)
Z
P (x)
dx, where the
Q(x)
degree (largest exponent) of P (x) is smaller than the degree of Q(x) then factor the denominator
as completely as possible and find the partial fraction decomposition of the rational expression.
Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get
term(s) in the decomposition according to the following table.
Partial Fractions : If integrating a rational expression involving polynomials,
Factor of Q(x)
ax + b
ax2 + bx + c
Term in P.F.D
A
ax + b
Ax + B
ax2 + bx + c
Factor is Q(x)
(ax + b)k
(ax2 + bx + c)k
Term in P.F.D
A1
A2
Ak
+
+ &middot;&middot;&middot; +
2
ax + b (ax + b)
(ax + b)k
A1 x + B1
Ak x + Bk
+ &middot;&middot;&middot; +
ax2 + bx + c
(ax2 + bx + c)k
Products and (some) Quotients of Trig Functions :
Z
For sinn (x) cosm (x) dx we have the following :
1. n odd. Strip 1 sine out and convert rest to cosines using sin2 (x) = 1 − cos2 (x), then use the
substitution u = cos(x).
2. m odd. Strip 1 cosine out and convert rest to sines using cos2 (x) = 1 − sin2 (x), then use
the substitution u = sin(x).
3. n and m both odd. Use either 1. or 2.
4. n and m both even. Use double angle and/or half angle formulas to reduce the integral into
a form that can be integrated.
Z
For tann (x) secm (x) dx we have the following :
1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using
tan2 (x) = sec2 (x) − 1, then use the substitution u = sec(x).
2. m even. Strip 2 secants out and convert rest to tangents using sec2 (x) = 1 + tan2 (x), then
use the substitution u = tan(x).
3. n odd and m even. Use either 1. or 2.
4. n even and m odd. Each integral will be dealt with differently.
3
3
Convert Example : cos6 (x) = cos2 (x) = 1 − sin2 (x)
&copy; Paul Dawkins - https://tutorial.math.lamar.edu
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