## Techniques of Integration

**Techniques of Integration** Assignment Help | **Techniques of Integration** Homework Help

# Techniques of Integration

There exist many functions whose integral cannot be found directly by applying the basic integration formulas. As such some techniques must be developed for integrating more complicated functions. Of many techniques, we shall discuss only the following:

(a) Integration by substitution.

(b) Integration by parts.

(c) Integration by partial fractions.

(fg)’ = f ‘ g + fg ‘

Integrating both sides with respect to x, we obtain

∫ f (fg)’ dx = ∫ f ' g dx + ∫ fg ' dx

i.e., fg = ∫ f ' g dx + ∫ fg ' dx

f (fg ' )dx = fg - ∫ f ' g dx

Let us define functions us and v as follows:

Now v = g’⇒ g = ∫ v dx. Therefore, from Eq. we get

∫ u v dx = u ∫ v dx - ∫ [du/dx . ∫ v dx] dx,

which is known as the integration by parts formula. This can be stated as follows:

Integral of the product of two functions is equal to the first functions times the integral of the second minus the integral of the product of the derivative of first and the integral of second.

N (x) / D (x) = Q(x) + R (x) / D (x),

where Q (x) is the quotient (a polynomial : easily integrable) and R(x) is the remainder whose degree is less than the degree of D (x).

The method of expressing N(x) / D (x) into partial fractions depends on factors of the denominator. It is a fact that every polynomial D (x) can be expressed as a product of linear and irreducible quadratic (one that can not be expressed as a product of two linear facotors with real coefficients) facotrs with real coefficients.

For more help in

(a) Integration by substitution.

(b) Integration by parts.

(c) Integration by partial fractions.

**(a) Integration by substitution.**Many a times an integrand is reduced to a standard form by making a suitable substitution. This is illustrated in the following examples.**(b) Integration by parts.**This technique is used to integrate the product of two functions. It is derived from the product rule of differentiation. We known that if f and g are two differentiable functions of x, then(fg)’ = f ‘ g + fg ‘

Integrating both sides with respect to x, we obtain

∫ f (fg)’ dx = ∫ f ' g dx + ∫ fg ' dx

i.e., fg = ∫ f ' g dx + ∫ fg ' dx

f (fg ' )dx = fg - ∫ f ' g dx

Let us define functions us and v as follows:

Now v = g’⇒ g = ∫ v dx. Therefore, from Eq. we get

∫ u v dx = u ∫ v dx - ∫ [du/dx . ∫ v dx] dx,

which is known as the integration by parts formula. This can be stated as follows:

Integral of the product of two functions is equal to the first functions times the integral of the second minus the integral of the product of the derivative of first and the integral of second.

**(c) Integration by partial fractions.**This method is used to integrate functions of the type N(x)/D(x), where both N (x) and D (x) are polynomials. Such functions are called rational functions. We may assume, without any loss of generality, that the numerator N (x) and denominator N(x) is less than the degree of D (x). For, if the degree of numerator were greater than or equal to the degree of denominator, we could use long division to divide N (x) so thatN (x) / D (x) = Q(x) + R (x) / D (x),

where Q (x) is the quotient (a polynomial : easily integrable) and R(x) is the remainder whose degree is less than the degree of D (x).

The method of expressing N(x) / D (x) into partial fractions depends on factors of the denominator. It is a fact that every polynomial D (x) can be expressed as a product of linear and irreducible quadratic (one that can not be expressed as a product of two linear facotors with real coefficients) facotrs with real coefficients.

For more help in

**Techniques of Integration**click the button below to submit your homework assignment