Calculus II: Integration techniques
1. Basic formulas
They come from the differentiation rules.
∫ k f(x) dx = k ∫ f(x) dx
∫ k dx = k x + const.
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
∫ xn dx = xn+1/(n+1) + const; with n ≠ -1
∫ sinxdx = - cos x + const.
∫ cosdx = sinx + const.
∫ sec2xdx = tanx + const.
∫ csc2xdx = -cotx + const.
∫ secx tanx dx = secx + const.
∫ cscx cotx dx = - cscx + const.
∫ ex dx = ex + const.
∫ ax dx = ax/ln a + const. with a >0 and ≠1.
∫ dx/x = ln |x| + const.
∫ tang x dx = - ln |cos x| + const.
∫ cot x dx = ln |sin x| + const.
∫ sec x dx = ln |sec x + tan x| + const.
∫ csc x dx = - ln |csc x + cot x| + const.
∫ dx/[a2 - x2]1/2 = arcsin (x/a) + const.
∫ dx/(a2 + x2)= (1/a)arctan (x/a) + const.
∫ dx/x[x2 - a2]1/2 = (1/a)arcsec(x/a) + const.
2. Change of variables
It makes the indefinite integral much easier to evaluate.
We must write the result in terms of the original variable
of integration, thereafter.
Example 1
I = ∫x2 dx/(x3 - 1)5
u = x3 - 1 → du = 3x2 dx, so
x2 dx = du/3.
Therefore
I = ∫du/3u5 = - (1/12)u-4 + const.
= - 1/12(x3 - 1)4 + const.
∫x2dx/(x3 - 1)5 =
- 1/12(x3 - 1)4 + const.
Example 2
I = ∫ x dx/[1 - x4]1/2
u = x2 → du = 2 x dx, so
x dx = du/2.
Therefore
I = (1/2)∫ du/[1 - u2]1/2 + const. =
(1/2)arcsin u + const. = (1/2)arcsin (x2) + const.
∫x dx/[1 - x4]1/2 =
(1/2)arcsin (x2) + const.
Example 3
I = ∫ dx/(x1/2 + x1/3)
The smallest number divisible by 2 and 3 is 6, so
x = u6 → 6 u5 du = dx, thus
I = ∫ 6 u5 du /(u3 + u2) =
6 ∫ u3 du /(u + 1)
We have the following polynomial division:
u3/(u + 1) = u2 - u + 1 - 1/(u + 1)
Therefore
I = 6 [∫ u2du - ∫ u du + ∫ du - ∫ du/(u + 1)]
= 6[u3/3 - u2/2 + u - ln|u + 1|]
= 6 [x1/2/3 - x1/3/2 + x1/6 - ln (x1/6 + 1)] + const.
∫dx/(x1/2 + x1/3) =
2x1/2 - 3x1/3 + 6x1/6 - 6ln (x1/6 + 1)] + const.
3. Integration by parts
The integration by parts is based on the inverse of
the derivative of product of two functions.
The basic formula:
∫u dv = uv - ∫v du
Example 1
I = ∫ x ex dx
u = x → du = dx
ex dx = dv → v = ex
so
I = ∫ u dv = uv - ∫v du = x ex - ∫ ex dx
= x ex - ex + const.
Example 2
I = ∫ ln x dx
u = ln x → du = dx/x
dx = dv → v = x
so
I = ∫ u dv = uv - ∫v du = x ln x - ∫ x dx /x
= x ln x - x + const.
4. Trigonometric integrals
sin and cos
The basic form:
∫ cosm x sinn x dx
m or n or both are odd:
Use
cos2 x + sin2 x = 1,
Example
I = ∫ cos3 x sin4 x dx
cos2 x + sin2 x = 1
d = sin x → du = cos x dx
I = (1/5)sin5x - (1/7)sin7x + const.
m and n both are even:
Use
cos2 x = (1/2)[1 + cos(2x)]
sin2 x = (1/2)[1 - cos(2x)]
Example
I = ∫cos4x sin2 x dx
= (1/16)x-(1/64)sin(4x)+(1/48)sin3(2x)+const.
sec and tan
∫ tanm x secn x dx
n even:
Use
u = tan x
Example
I = ∫ tan2 x sec4 x dx
I = (1/5)tan5x + (1/3)tan3x + const.
m is odd:
Use
tan2 x += sec2 x - 1,
u = sec x
Example
I = ∫ tan x sec6 x dx
= (1/6)sec6x + const.
c. m is even and n odd:
Use
tan2 x += sec2 x - 1,
Example
I = ∫ tan2 x sec3 x dx
= tan x sec3 /4 - tan x sec x /8 - (1/8) ln|sec x + tan x| + const.
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Exercices:
Changing variables:
1. ∫ dx/(1 + ex)
2. ∫ sec x dx
By parts:
3. ∫ x2e-x dx
4. ∫ sin x e2x dx
5. ∫ x e2x dx
6. ∫ cosnx dx
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