Let us consider We observe that the cos x is the derivative function of sin x or we can say that sin x is an antiderivative (or an integral) of cos x.
Some Important Integrals
Some Properties of Indefinite Integrals
Methods of Integration
1. Integration by Substitution
The given integral ∫f(x)dx can be transformed into another form by changing the independent variable x to t by substituting x = g(t) Consider, I = ∫f(x)dx
Put, x = g(t) so that dx / dt = g'(t)
dx = g'(t)dt
I = ∫f(x)dx = ∫f(g(t)) g'(t)dt
Example 1:
Integrate the following function,
Solution:
Derivative of √x is = 1 / 2√x
We use the substitution √x = t
So, that = dt ⇒ dx = 2t dt
Thus,
= 2∫tan^{4} t sec^{2} t dt
Again we make another substitution tan t = u
sec2 t dt = dx
= 2∫ x4 dx = 2u^{5}/ 5 + c
=
[since u = tan t]
=
Note:
2. Integration using Trigonometric Identities
The following example can be solved by using trigonometric Identities
Example 2:
∫cos2x dx
Solution:
We know, cos 2x = 2 cos^{2} x – 1
=
=
Integrals of Some Particular Functions
To find the integral of type , where p,q,a,b,c are constants we need to find real numbers A, B such that
To determine A and B we equate from both sides the coefficients of x and the constant terms.
3. Integration by Partial Fractions
The following example can be solved by using partial fraction method
Example 3:
Find
Solution:
=
4. Integration by Parts
This can be done by using the given formula
f(x)g(x)dx = f(x)∫g(x)dx – ∫[∫g(x)dx] f'(x)dx
In integration by parts method proper choice of first and second function:
The first function is the word ILATE
I → Inverse Trigonometric
L → Logarithmic
A → Algebraic
T → Trigonometric
E → Exponential
Example 4:
Find the value of ∫x cos x dx
Solution:
Put f(x) = x (first function) and g(x) = cos x (second function)
= x sin x – ∫sin x dx
= x sin x + cos x + c
Integrals of Some Particular Functions
A real valued function f(x) is continuous on [a, b] then is called definite integral. Geometrically it gives area of finite regions.
Properties of Definite Integrals
Example 5:
Find the value of
Solution:
Let, I =
Also, I = = =
Solving Indefinite Integrals Using Functions
There are two functions which can be used to solve improper indefinite integrals. They are given as
A. Gamma Function
Gamma Function is defined by:
Properties
B. β – Function (Beta Function)
It is defined as β(m, n) =
Properties
Example 6:
Find the value of
Solution:
2m  1 = 1 / 2
⇒ m = 3 / 4
⇒ 2n  1 = 1 / 2
⇒ n = 1 / 4
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