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, (∫u dv = uv - ∫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 ( ∫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.

Special treatment of the ln function: It's easy to forget that ln(u) = 2 ln(sqrt(u)), and sometimes the latter appears easier to integrate, for example, if u=x^2+a^2.  Similarly, if the ln function appears in the solution of an integral, it is important to remember that, for example, ln(u) and ln(3u) differ by a constant.

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.
Trig substitution is a technique that helps with many of these.

 . . . . . . write up the integral of (x+2x^-1/2)^1/2dx, so we make the substitution
x=u^4/3, which means
dx = (4/3)u^1/3du
Substituting, we get (4/3)(u^4/3+2u^-2/3)^1/2u^1/3du.
you can distribute the u^1/3 into the square root by squaring it... so we get
(4/3)(u^6/3+2u^0/3)^1/2du
(4/3)(u²+2)^1/2du
which is solved using sqrt(a²+x²), above, and then substituting back . . . . . .

 . . . . . . add this as a separate page, proving the integral of 1/sqrt(a^2+x^2)...
The derivative with respect to x of ln(u) is (1/u)(du/dx). So the derivative of ln(x + sqrt(1+x^2)) is (1 + x/sqrt(1+x^2))/(x + sqrt(1+x^2)), which simplifies to 1/sqrt(1 + x^2) using the fact that when a/b = c/d, it follows that a/b = (a+c)/(b+d) as well.

 . . . . . . add the following factoid:

int ln(1+x^2)dx = int 2 ln(sqrt(1+x^2))dx
let x = tan u, so dx = sec^2 u du, and sqrt(1+x^2)=sec u, so
= int 2 (sec^2 u)(ln(sec u)) du
= (tan u) (ln(sec^2 u)) - 2(tan u + u) (by parts)
= (tan u) (ln(1+tan^2 u)) - 2(tan u + u)
= x (ln(1+x^2) - 2x + 2 arctan x

more generally, int ln(x^2+a^2) = x ln(x^2 + a^2) - 2x + 2a arctan(x/a)  

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

∫ sin(ax) sin(bx) dx =
(1/2) ∫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

∫ sin(ax) cos(bx) dx =
(1/2) ∫ 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

∫ cos(ax) cos(bx) dx =
(1/2) ∫ 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 = ∫ (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 = ∫ a² cos x dx/(a sin x + b cos x)

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

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

Now,

I₁ - I₃ = a ∫ (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 ∫ (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 = ∫ (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 = ∫ a² sin x dx/(a sin x + b cos x)

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

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

Now,

I₂ - I₃ = -b ∫ (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 ∫ (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 = ∫ (1-sin x)/(1+sin x) dx

∫(1-sin x)/(1+sin x) dx
= ∫(1-sin x)²/(1-sin²x) dx
= ∫(1 - 2 sin x + sin²x)/cos²x dx
= ∫sec²x dx  - ∫2 sin x/cos²x dx + ∫tan²x dx
= ∫sec²x dx  - ∫2 sin x/cos²x dx + ∫(sec²x -1) dx
= tan x - 2 sec x + (tan(x)-x) + C
= 2 tan(x) - 2 sec(x) - x + C

I = ∫ 1/(1+x²)²

Rewrite as ∫(1/2)(1+x²)/(1+x²)² dx + ∫(1/2)(1-x²)/(1+x²)² dx
The first integral is ∫(1/2)(1/(1+x²)) dx, which is (1/2)arctan(x)
The second integral is ∫(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))

I = ∫ (x²-1)/(x⁴+1) dx

Factor x⁴+1 as (x²-sqrt(2)x+1)(x²+sqrt(2)x+1)
Then let x²-1 = (Ax+B)(x²+sqrt(2)x+1) + (Cx+D)(x²-sqrt(2)x+1)
 = (A+C)x³ + (sqrt(2)A+B-sqrt(2)C+D)x² + (A+sqrt(2)B+C-sqrt(2)D)x + (B+D), which by equating coefficients, gives us the following system of linear equations:

A + C = 0
sqrt(2)A + B - sqrt(2)C + D = 1
A + sqrt(2)B + C - sqrt(2)D = 0
B + D = -1

Solving this system gives us A=sqrt(2)/2, B=-1/2, C=-sqrt(2)/2, D=-1/2, so

∫(x²-1)/(x⁴+1) dx = (1/2)∫((sqrt(2)x-1)/(x^2-sqrt(2)x+1) + (-sqrt(2)x-1)/(x^2+sqrt(2)x+1)) dx

= (1/sqrt(8))(ln(x²-sqrt(2)x+1)-ln(x²+sqrt(2)x+1))

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=-π/4, this becomes
   tan x-π/4 = (tan x - 1)/(tan x + 1)

Integration by Parts

Weierstrass t-substitution is a clever trig substitution that lets you solve the kind of integrals that naturally come up in the polar form of certain functions. The substitution is t=tan(x/2).

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