mathematics MTH

Integration by parts – formula, proof, derivation, examples and FAQs


WHAT IS INTEGRATION BY PARTS?

Integration by parts is used to integrate the product of two or more functions. The two functions to be integrated f(x) and g(x) are of the form

f(x).g(x). Thus, it can be called a product rule of integration. Among the two functions, the first function f(x) is selected such that its derivative formula exists, and the second function g(x) is chosen such that an integral of such a function exists.

INTEGRATION BY PARTS FORMULA DERIVATION

The proof of integration by parts can be obtained from the formula of the derivative of the product of two functions. For the two functions f(x) and g(x), the derivative of the product of these two functions is equal to the sum of the derivatives of the first functions multiplied with the second function, and the derivative of the second function multiplied by the first function.

Let us derive the integration by parts formula using the product rule of differentiation. Consider two functions u and v. Let their product be y. i.e., y = uv. Applying the product rule of differentiation, we get

d/dx (uv) = u (dv/dx) + v (du/dx)

We will rearrange the terms here.

u (dv/dx) = d/dx (uv) – v (du/dx)

Integrating on both sides with respect to x,

∫ u (dv/dx) (dx) = ∫ d/dx (uv) dx – ∫ v (du/dx) dx

By cancelling the terms,

∫ u dv = uv – ∫ v du

Hence the integration by parts formula is derived.

Visualizing Integration by Parts

Applications Of Integration By Parts
The application of this formula for integration by parts is for functions or expressions for which the formulas of integration do not exist. Here we try to include this formula of integration by parts and try to derive the integral. For logarithmic functions and for inverse trigonometry functions there are no integral answers. Let us try to solve and find the integration of log x and tan-1x.

Integration of Logarithmic Function
∫ logx.dx = ∫ logx.1.dx

= logx. ∫1.dx – ∫ ((logx)’.∫ 1.dx).dx

=logx.x -∫ (1/x .x).dx

=xlogx – ∫ 1.dx

=x logx – x + C

Integration of Inverse Trigonometric Function
∫ tan-1x.dx = ∫tan-1x.1.dx

= tan-1x.∫1.dx – ∫((tan-1x)’.∫ 1.dx).dx

= tan-1x. x – ∫(1/(1 + x2).x).dx

= x. tan-1x – ∫ 2x/(2(1 + x2)).dx

= x. tan-1x – ½.log(1 + x2) + C

Formulas Related to Integration by Parts

The following formulas have been derived from the integration by parts formula and are helpful in the process of integrations of various algebraic expressions.

∫ ex(f(x) + f'(x)).dx = exf(x) + C
∫√(x2 + a2).dx = ½ . x.√(x2 + a2)+ a2/2. log|x + √(x2 + a2)| C
∫√(x2 – a2).dx =½ . x.√(x2 – a2) – a2/2. log|x +√(x2 – a2) | C
∫√(a2 – x2).dx = ½ . x.√(a2 – x2) + a2/2. sin-1 x/a + C
Example

Find ∫xe-x dx

Integrating by parts (with v = x and du/dx = e-x), we get:

-xe-x – ∫-e-x dx (since ∫e-x dx = -e-x)

= -xe-x – e-x + constant

We can also sometimes use integration by parts when we want to integrate a function that cannot be split into the product of two things. The trick we use in such circumstances is to multiply by 1 and take du/dx = 1.

EXAMPLE

Find ∫ ln x dx

To integrate this, we use a trick, rewrite the integrand (the expression we are integrating) as 1.lnx . We then let v = ln x and du/dx = 1 .

Hence ∫ ln x dx = x ln x – ∫ x (1/x) dx
= x lnx – ∫ dx
= x lnx – x + constant
Also check https://revisionmaths.com/advanced-level-maths-revision/pure-maths/calculus/integration-part

Solved Examples On Integration By Parts

Example 1: Find the integral of x2ex by using the integration by parts formula.

Solution:

Using LIATE, u = x2 and dv = ex dx.

Then, du = 2x dx, v = ∫ ex dx = ex.

Using one of the integration by parts formulas,

∫ u dv = uv – ∫ v du

∫ x2 ex dx = x2 ex – ∫ ex (2x) dx

= x2 ex – 2 ∫ x ex dx

Applying integration by parts formula again to evaluate ∫ x ex dx,

∫ x2 ex dx = x2 ex – 2 (x ex – ∫ ex dx) = x2 ex – 2 x ex + 2 ex + C

= ex (x2- 2 x + 2)+ C

Answer: ∫ x2 ex dx = = ex (x2- 2 x + 2)+ C

Example 2: Find the integral of x sin2x, by using integration by parts formula.

Solution:

To find the integration of the given expression we use the integration by parts formula: ∫ uv.dx = u∫ v.dx -∫( u’ ∫ v.dx).dx

Here u = x, and v = Sin2x

∫x sin2x. dx

=x∫sin2xdx – d/dx. x.∫ sin2xdx. dx

=x. -cos2x/2 – ∫(1.-cos2x/2). dx

=-cos2x/2. dx + 1/2 cos2xdx

=-xcos2x/2 + sin2x/4 + C

Answer: Thus ∫x sin2x dx = -x cos2x/2 +sin 2x/4+ C

Example 3: Evaluate the integral ∫ x ln x dx using integration by parts.

Solution:

First Method:

Using LIATE, u = ln x and v = x.

Using one of the formulas of integration by parts,

∫ uv dx = u ∫ v dx – ∫ (u’ ∫ v dx) dx

∫ x ln x dx = ln x ∫ x dx – ∫ (1/x) (∫ x dx) dxU

= ln x (x2/2) – ∫ (1/x) (x2/2) dx

= (x2 ln x)/2 – (1/2) ∫ x dx

= (x2 ln x) / 2 – (1/2) (x2/2) + C

= (x2 ln x) /2 – (x2 / 4) + C

=(x2/4)(2 ln x -1) + C

Second Method:

Using LIATE, u = ln x and dv = x dx.

Then du = (1/x) dx and v = ∫ x dx = x2/2

Using one of the formulas of integration by parts,

∫ u dv = uv – ∫ v du

∫ x ln x dx = ln x (x2/2) – ∫ (x2/2) (1/x) dx

= (x2 ln x)/2 – (1/2) ∫ x dx

= (x2 ln x) / 2 – (1/2) (x2/2) + C

= (x2 ln x) /2 – (x2 / 4) + C

= (x2/4)(2 ln x -1) + C

Practice Questions on Integration by Parts

Q 1. Find ∫ sin-1x. dx using integration by parts.

x sin-1x -√(x2 -1 )+ C.
x sin, -1, x -√(x, 2, -1, , )+ C.

x sin-1x + √(1-x2) + C.
x sin, -1, x + √(1-x, 2, ) + C.

x sin-1x -√(1-x2) + C.
x sin, -1, x -√(1-x, 2, ) + C.

x sin-1x -√(x2 -1 ) + C.
x sin, -1, x -√(x, 2, -1, , ) + C.
Q 2. For which of the following integral expressions we need to apply Integration by Parts formula.

∫ log sinx dx
∫ log sinx dx

∫ x ex . dx
∫ x e, x , . dx

∫ cosec2 x. dx
∫ cosec, 2 , x. dx

∫ tan x. dx
∫ tan x. dx

Oluwamuyide Peter

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