Calculus II: Methods of integrating
1. Change of variables
The change of variables method is the most powerful
thechnique to calculaute the primitive of a function.
With this method, we change the expression of the integrand
to obtain another more simple to integrate by changing the
independent variable into a new variable.
2. Examples
2.1. Example 1: power function
1. We do not know how to integrate the following
function:
f(x) = (3x + 1)3 , by using x as the differential
element. Therefore by the following change:
3x + 1 = u , so du = 3 dx, we write:
∫ (3x + 1)3 dx = ∫ u3 3 du =
3 (1/4) u4 + cst = (3/4) (x + 1 )4 + cst.
2.2. Example 2: polynomial function
f(x) = (4x3 + 2)(x4 + 2 x + 5)
Let u = x4 + 2 x + 5, so
du = (4x3 + 2) dx, then:
∫ (4x3 + 2)(x4 + 2 x + 5) dx =
∫ d du = (1/2) u2 = (1/2) (x4 + 2 x + 5)2 + cst.
2.3. Example 3: Involving ln
f(x) = 3/(2 x + 1)
Let u = 2 x + 1, so
du = 2 dx, then:
∫ 3/(2 x + 1) dx = 3 ∫ dx/(2 x + 1) =
3 ∫ (du/2)/u = (3/2) ∫ du/u
= (3/2) ln |u| + cst = (3/2) ln |2x + 1| + cst
2.4. Example 4: Rational functions
f(x) = (x2 + x + 3)/( x + 1)
For the rational function, if the degree of the numerator is
greater that the degree of the denominator, we perform
the division of polynomials. So
(x2 + x + 3)/( x + 1) = x + 3/(x + 1)
Therefore
∫ (x2 + x + 3)/( x + 1) dx = ∫ [x + 3/(x + 1)] dx =
∫ x dx + ∫ 3 dx /(x + 1) = (1/2) x2 + 3 ∫ dx /(x + 1) =
(1/2) x2 + 3 ln|x + 1| + cst.
3. Some other useful types of primitive functions
3.1. Type 1: ∫ ax dx
Let u = ax, so
x = ln u /ln a, and dx = d(ln u /ln a) = = du/(u ln a)
Therefore
∫ ax dx = ∫ u du/(u ln a) =
(1/ln a) ∫ du = (1/ln a) u + cst. =
ax /ln a + cst.
a > 0 and ≠ 1 of course.
∫ ax dx = ax/ln a + cst
a > 0 and ≠ 1.
3.2. Type 2: ∫ ex dx
Since ln(e) = 1, we have
∫ ex dx = ex + cst
3.3. Type 3: Trigonometric fontions
∫ cos x dx = sin x + cst
∫ sin x dx = - cos x + cst
∫ tan x dx = ln |1/cos x| + cst
∫ cot x dx = ln |sin x| + cst
3.4. Type 4: Iinverse trigonometric fontions
∫ dx/(a2 - x2)1/2 = arcsin(x/a) + cst
∫ dx/(a2 + x2) = (1/a) arctan(x/a) + cst
∫ dx/x(x2 - a2 )1/2 =
(1/a) arcsec(x/a) + cst
∫ cot x dx = ln |sin x| + cst
4. Exercise
|
