table_of_integrals

8/7/2019 Table_of_Integrals

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MATH 2170: Table of Derivatives and Integrals

1. Basic Properties of Derivatives:

d

dx(cf(x)) = c f(x), c is any constant (f(x)

g(x)) = f(x)

g(x)

d

dx(xn) = nxn1, n is any number

d

dx(C) = 0, C is any constant

(f g) = fg + gf (Product Rule) (f

g) =

fg gfg2

(Quotient Rule)

d

dx(f(g(x)) = f(g(x)) g(x) (Chain Rule)

d

dx(eg(x)) = g(x) eg(x)

d

dx(ln(g(x)) =

g(x)

g(x)

2. Common Derivatives:

Trig Functions:

d

dx(sinx) = cosx

d

dx(cosx) = sinx d

dx(tanx) = sec2x

d

dx(secx) = secx tanx

d

dx(cscx) =

cscx cotx

d

dx(cotx) =

csc2x

Inverse Trig Functions:

d

dx(sin1x) =

11 x2

d

dx(cos1x) = 1

1 x2d

dx(tan1x) =

1

1 + x2

d

dx(sec1x) =

1

x

x2 1d

dx(csc1x) = 1

x

x2 1d

dx(cot1x) = 1

1 + x2

Exponential/Logarithmic Functions/ Polynomials

d

dx(ax) = axln(a)

d

dx(ex) = ex

d

dx(ln(x)) =

1

x, x > 0

d

dx(loga(x)) =

1

x ln a, x > 0

d

dx(xn) = nxn1


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Hyperbolic Trig Functions

d

dx(sinh x) = cosh x

d

dx(cosh x) = sinh x

d

dx(tanh x) = sech2xd

dx (sech x) = sech x tanh xd

dx(cschx) = cschx cothx d

dx(coth x) = csch2x

3. Basic properties/Formulas/Rules of Integration:

cf(x)dx = c

f(x)dx, c is any constant

(f(x) g(x))dx =

f(x)dx

g(x)dx

ba

f(x)dx = F(x)|ba = F(b) F(a) where F(x) =

f(x)dx

ba

cf(x)dx = c

ba

f(x)dx, c is a constant

b

a (f(x) g(x))dx = b

a f(x)dx b

a g(x)dx

aa

f(x) = 0;

ba

f(x) = ab

(f(x);

ba

f(x)dx =

ca

f(x)dx +

bc

f(x)dx

If f(x) 0 on a x b then ba

f(x)dx 0

If f(x) g(x) on a x b then ba

f(x)dx ba

g(x)dx


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4. Common Integrals:

Polynomials

dx=

x+

c kdx=

kx+

c

xndx =

1

n + 1xn+1 + c, n = 1;

x1dx = ln | x | +c;

1

xdx = ln | x | +c;

xndx =

1

n + 1 xn+1 + c, n = 1

1

ax + bdx =

1

aln | ax + b | + c;

x

p

q dx =1

(pq

+ 1)x

p

q+1 + c = (

q

p + q)x(

p+q

q) + c

Trig Functionscos xdx = sin x + c;

sin xdx = cos x + c; sec2 xdx = tan x + c

sec x tan x dx = sec x + c;

csc x cot xdx = csc x + c; csc2 x = cot x + c

sec x dx = ln | sec x + tan x | +c;

sec3 xdx =

1

2(sec x tan x + ln | sec x + tan x |) + c

csc x dx = ln |csc x

cot x

|+c; csc

3 xdx =1

2

(

csc x cot x + ln

|csc x

cot x

|) + c

Exponential/Logarithmic Functionsexdx = ex + c;

ax = a

x

ln a+ c;

ln xdx = x ln(x) x + c

eax sin(bx)dx =

eax

a2 + b2(a sin(bx) b cos(bx)) + c;

xexdx = (x 1)ex + c

eax cos(bx)dx =

eax

a2 + b2(a cos(bx) + b sin(bx)) + c;

1

x ln xdx = ln | ln x | +c

Inverse Trig Functions1

a2 x2 dx = sin1(

x

a) + c;

sin1 xdx = x sin1 x +

1 x2 + c

1

a2 + x2dx =

1

atan1(

x

a) + c;

tan1 xdx = x tan1 x 1

2ln(1 + x2) + c

1

x

x2

a2

dx =1

asec1(

x

a) + c;

cos1 xdx = x cos1 x1 x2 + c


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Hyperbolic Trig Functionssinh x dx = cosh x + c;

cosh xdx = sinh x + c;

sech2x dx = tanh x + c

sechx tanh xdx = sechx + c; cschx coth x dx = cschx + c

tanh xdx = ln(cosh x) + c;

csh2xdx = coth x + c

sechx dx = tan1 | sinh x | +c

Miscellaneous

1a2 x2 dx =

1

2a ln |x + a

x a | +c; 1

x2 a2 dx =1

2a ln |x

a

x + a | +c

a2 + x2dx =x

2

a2 + x2 +

a2

2ln | x +

a2 + x2 | +c

x2 a2dx = x

2

x2 a2 a

2

2ln | x +

x2 a2 | +c

a2 x2 dx = x

2

a2 x2 + a

2

2sin1(

x

a) + c

2ax x2 dx = (x a)

2

2ax x2 + a

2

2cos1(

a xa

) + c

5. Standard Integration Techniques :

Given below are some of the standard integration techniques you would have learned in yourCalculus courses.

u Substitution

Givenba

f(g(x))g(x)dx, use the substitution u = g(x) and convert the given integral into

the integral

ba

f(g(x))g(x)dx =

g(b)g(a)

f(u)du.

Integration by parts

The standard formula to remember for integration by parts is given below;udv = uv

vdu

ba

udv = uv |ba ba

vdu

The task is to cleverly choose u and dv and then compute du by differentiating u and compute

v by using the fact that v =

dv.


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Trig substitutions

If the integral contains the following roots try using the corresponding substitution given be-low and use trig formulas.

a2 b2x2 = x =a

b sin and use cos2 = 1 sin2

b2x2 a2 = x = a

bsec and use tan2 = sec2 1

a2 + b2x2 = x = a

btan and use sec2 = 1 + tan2

Partial Fractions

If integrating the rational expression

P(x)Q(x)

dx, first check whether the degree of the polynomial

P(x) is less than degree of polynomial in Q(x). If not, using long division factor out till weget an degree of polynomial of P(x) less than degree of Q(x).

Step 1 : Factorize the denominator polynomial Q(x). Step 2 : Using rules given in table below to write out the partial fraction decomposition

(P.F.D) of rational expression.

Step 3: Having obtained the partial fraction decomposition, integrate each term in thedecomposition separately and add all the integrals to get the integral of the rationalexpression.

Factor in Q(x Term in P.F.D

ax + bA

ax + b

(ax + b)kA1

ax + b+

A2

(ax + b)2+ . . . +

Ak

(ax + b)k

ax2 + bx + cAx + B

ax2 + bx + c

(ax2 + bx + c)kA1x + B1

ax2 + bx + c+ . . . +

Akx + bk(ax2 + bx + c)k

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