Tables of Integrals

On this page you will find a list of many of the most common integrals, organized by how "hard" they are.  "Hard" means it gets "messier" when you integrate it, and "easy" means it doesn't.  Moreover, the "hardest" integrals often have "easy" derivatives, and vice-versa.  So when you do integration by parts, ( ò u dv = uv -  ò v du) you should pick dv as the easier, and u as the harder of the two functions to integrate, because you'll have to integrate dv and differentiate u.  You can remember the order of the categories because they spell "ILATE".  For dv, pick the latest one you can in that list.

I try to list the standard "integration by parts" integrals ( ò u dv ) together with the part, dv, i.e. latest in the "ILATE" list.  So you'll find x cos(x) together with cos(x), for example, because x is algebraic (A), and cos(x) is trigonometric (T), which comes later in "ILATE".

Constants: I don't show the constant of integration, and in fact, the integrals shown may differ by a constant from what you would expect to see, and this is not wrong.  For example, the integral of x/(ax+b) is shown as (1/a²)(ax + b - b ln|ax + b|), because it's "nice" to see every "x" as part of "ax+b".  This, despite the method described giving you (1/a²)(ax - b ln|ax + b|).  These two integrals differ only by a constant, so they are both right.

I: inverse trigonometric functions: arctan(x), arcsec(x), etc. 

L: logarithmic functions: ln(x), log₂(x), etc. 

A: algebraic functions: x², 3x⁵⁰, etc. 

Special case: 1/x, more generally, du/u:

T: trigonometric functions: sin(x), tan(x), etc. 

E: exponential functions: e^x, 13^x, etc. 

"Other"

Functions involving (a²+x²) or (a²-x²) or its square root:

Proofs of Selected Integrals

From the identity cos(a) - cos(b) = -2 sin((a-b)/2) sin((a+b)/2), which is proved here, we get:

(1/2) (cos(px) - cos(qx)) = sin((q-p)x/2) sin((q+p)x/2)  

Let a=(q-p)/2, and let b=(q+p)/2; then a+b=q and a-b=-p, and remembering cos(-x)=cos(x),

(1/2) (cos((a-b)x) - cos((a+b)x)) = sin(ax) sin(bx)

So the integral, above, is equal to

ò sin(ax) sin(bx) dx =
(1/2) ò cos((a-b)x) - cos((a+b)x)  =
(1/2)(sin(a-b)x/(a-b) - sin(a+b)x/(a+b)) + C

From the identity sin(a) + sin(b) = 2 cos((a-b)/2) sin((a+b)/2), which is proved here, we get:

(1/2) (sin(qx) + sin(px)) = sin((p+q)x/2) cos((p-q)x/2)  

Let a=(p+q)/2, and let b=(p-q)/2; then a+b=p and a-b=q, and therefore,

(1/2) (sin((a-b)x) + sin((a+b)x)) = sin(ax) cos(bx)

So the integral, above, is equal to

ò sin(ax) cos(bx) dx =
(1/2) ò sin((a-b)x) + sin((a+b)x)  =
-(1/2)(cos(a-b)x/(a-b) + cos(a+b)x/(a+b)) + C

From the identity cos(a) + cos(b) = 2 cos((a-b)/2) cos((a+b)/2), which is proved here, we get:

(1/2) (cos(px) + cos(qx)) = cos((p+q)x/2) cos((p-q)x/2)  

Let a=(p+q)/2, and let b=(p-q)/2; then a+b=p and a-b=q, and so,

(1/2) (cos((a+b)x) + cos((a-b)x)) = cos(ax) cos(bx)

So the integral, above, is equal to

ò cos(ax) cos(bx) dx =
(1/2) ò cos((a-b)x) + cos((a+b)x)  =
(1/2)(sin(a-b)x/(a-b) + sin(a+b)x/(a+b)) + C

(1/2)(sin(a-b)x/(a-b) + sin(a+b)x/(a+b))

I = ò(a²+b²) cos x dx / (a sin x + b cos x)

The approach is to break this down into the sum of two integrals, I₁ and I₂, and introduce a third integral, I₃, such that I₂+I₃ and I₁-I₃ are both easy to do, and so the sum of these two integrals is I₁+I₂, and that's our answer.

Let I₁ = ò a² cos x dx/(a sin x + b cos x)

Let I₂ = ò b² cos x dx/(a sin x + b cos x)

Let I₃ = ò ab sin x dx/(a sin x + b cos x)

Now,

I₁ - I₃ = a ò (a cos x - b sin x) dx/(a sin x + b cos x)
         = a ln | a sin x + b cos x | + C

I₂ + I₃ = b ò (a sin x + b cos x) dx/(a sin x + b cos x)
          = b x + D

Now add up these equations, and you get

I₁ + I₂ = I =  b x + a ln | a sin x + b cos x | + E

I = ò(a²+b²) sin x dx / (a sin x + b cos x)

The approach is to break this down into the sum of two integrals, I₁ and I₂, and introduce a third integral, I₃, such that I₁+I₃ and I₂-I₃ are both easy to do, and so the sum of these two integrals is I₁+I₂, and that's our answer.

Let I₁ = ò a² sin x dx/(a sin x + b cos x)

Let I₂ = ò b² sin x dx/(a sin x + b cos x)

Let I₃ = ò ab cos x dx/(a sin x + b cos x)

Now,

I₂ - I₃ = -b ò (a cos x - b sin x) dx/(a sin x + b cos x)
         = -b ln | a sin x + b cos x | + C

I₁ + I₃ = a ò (a sin x + b cos x) dx/(a sin x + b cos x)
          = a x + D

Now add up these equations, and you get

I₁ + I₂ = I = a x - b ln | a sin x + b cos x | + E

I = ò(1-sin x)/(1+sin x) dx

ò(1-sin x)/(1+sin x) dx
= ò(1-sin x)²/(1-sin²x) dx
= ò(1 - 2 sin x + sin²x)/cos²x dx
= òsec²x dx  - ò2 sin x/cos²x dx + òtan²x dx
= òsec²x dx  - ò2 sin x/cos²x dx + ò(sec²x -1) dx
= tan x - 2 sec x + (tan(x)-x) + C
= 2 tan(x) - 2 sec(x) - x + C

I = ò1/(1+x²)²

Rewrite as ò(1/2)(1+x²)/(1+x²)² dx + ò(1/2)(1-x²)/(1+x²)² dx
The first integral is ò(1/2)(1/(1+x²)) dx, which is (1/2)arctan(x)
The second integral is ò(1/2)(1-x²)(1+x²)^-2 dx, and the integral of this is (1/2)(x)(1+x²)^-1
So the answer is (1/2)(x/(1+x²) + arctan(x))

Internet References

M.G. Worster, Practice Integrals 

Related Pages in this Website

How to integrate sqrt(tan(x))

Trig Equivalences, which proves these identities: 
   (1-sin 2x)/(1+sin 2x) = (1 - tan(x))²/(1 + tan(x))² 
   1-cos(2u) = 2sin²u
   tan x+y = (tan x + tan y)/(1 - tan x tan y), and when y=-p/4, this becomes
   tan x-p/4 = (tan x - 1)/(tan x + 1)

Integration by Parts

The webmaster and author of the Math Help site is Graeme McRae.
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