Integration by Parts

Integration by Parts or Partial Integration, is a technique used in calculus to evaluate the integral of a product of two functions. The formula for partial integration is given by:

∫ u dv = uv - ∫ v du

Where u and v are differentiable functions of x. This formula allows us to simplify the integral of a product by breaking it down into two simpler integrals.

Suppose we have two functions f(x) and g(x) and we have to find the integration of their product i.e., ∫ f(x).g(x) dx where it is not possible to further solve the product of this product f(x).g(x). 

This integration is achieved using the formula:

∫ f(x).g(x) dx = f(x) ∫ g(x) d(x) - ∫ [f'(x) {∫g(x) dx} dx] dx + c

Where f'(x) is the first differentiation of f(x).

This formula is read as:

Integration of the First Function multiplied by the Second Function is equal to (First Function) multiplied by (Integration of Second Function) - Integration of (Differentiation of First Function multiplied by Integration of Second Function).

Note: Integration by part concept was first proposed by the famous Brook Taylor in his book in 1715. He wrote that we can find the integration of the product of two functions whose differentiation formulas exist.

Read More about Method of Integration.

Integration By Parts Formula

Integration by parts formula is the formula that helps us to achieve the integration of the product of two or more functions. Suppose we have to integrate the product of two functions as

∫u.v dx 

where u and v are the functions of x, then this can be achieved using,

∫u.v dx = u ∫ v d(x) - ∫ [u' {∫v dx} dx] dx + c

The order to choose the First function and the Second function is very important and the concept used in most of the cases to find the first function and the second function is ILATE concept.

Derivation of Integration By Parts Formula

Integration By Parts Formula is derived using the product rule of differentiation. Suppose we have two functions u and v and x then the derivative of their product is achieved using the formula,

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

Now to derive the integration by parts formula using the product rule of differentiation.

Rearranging the terms

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

Integrating both sides with respect to x,

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

simplifying,

∫ u dv = uv - ∫ v du

Thus, the integration by parts formula is derived.

ILATE Rule

The ILATE rule tells us about how to choose the first function and the second function while solving the integration of the product of two functions. Suppose we have two functions of x u and v and we have to find the integration of their product then we choose the first function and the by ILATE rule.

The ILATE full form is discussed in the image below,

The ILATE rules give us the hierarchy of taking the first function, i.e. if in the given product of the function, one function is a Logarithmic function and another function is a Trigonometric function. Now we take the Logarithmic function as the first function as it comes above in the hierarchy of the ILATE rule similarly, we choose the first and second functions accordingly.

NOTE: It is not always appropriate to use the ILATE rule sometimes other rules are also used to find the first function and the second function.

How to  Find Integration by Part?

Integration by part is used to find the integration of the product of two functions. We can achieve this using the steps discussed below,

Suppose we have to simplify ∫uv dx

Step 1: Choose the first and the second function according to the ILATE rule. Suppose we take u as the first function and v as the second function.

Step 2: Differentiate u(x) with respect to x that is, Evaluate du/dx.

Step 3: Integrate v(x) with respect to x that is, Evaluate ∫v dx.

Use the results obtained in Step 1 and Step 2 in the formula,

∫uv dx = u∫v dx − ∫((du/dx)∫v dx) dx

Step 4: Simplify the above formula to get the required integration.

Tabular Integration by Parts

Tabular integration, also known as the DI method of integration, is an alternative technique for evaluating integrals that involve repeated application of integration by parts.

This method is particularly useful when dealing with integrals where the product of functions can be integrated multiple times to reach a simple result.

Here's how the tabular method works:

1. Begin by writing down the functions involved in the integral in two columns: one for the function to differentiate (u) and another for the function to integrate (dv).
   • Start with the function to integrate (dv) on the left column and the function to differentiate (u) on the right column.
2. Continue differentiating the function in the u column until you reach zero or a constant. At each step, integrate the function in the dv column until you reach a point where further integration is not necessary.
3. Multiply the terms diagonally and alternate the signs (+ and -) for each term. Sum up these products to find the result of the integration.

Here's an example to illustrate the tabular integration method:

Let's evaluate the integral ∫x sin(x) dx.

• Step 1: Create a table with two columns for u (function to differentiate) and dv (function to integrate):

• Step 2: Differentiate the function in the u column and integrate the function in the dv column:

• Step 3: Multiply the terms diagonally and alternate the signs:

(x)(-cos(x)) - (1)(-sin(x)) + (0)(cos(x)) = -x cos(x) + sin(x)

So, the result of the integral ∫x sin(x) dx is -xcos(x) + sin(x).

The tabular integration method is especially useful when dealing with integrals that involve functions that repeat upon differentiation or integration, allowing for a systematic and organized approach to finding the antiderivative.

Applications of Integration by Parts

Integration by Parts has various applications in integral calculus it is used to find the integration of the function where normal integration techniques fail. We can easily find the integration of inverse and logarithmic functions using the integration by parts concept.

We will find the Integration of the Logarithmic function and Arctan function using integration by part rule,

Integration of Logarithmic Function (log x)

Integration of Inverse Logarithmic Function (log x) is achieved using Integration by part formula. The integration is discussed below,

∫ logx.dx = ∫ logx.1.dx

Taking log x as the first function and 1 as the second function.

Using ∫u.v dx = u ∫ v d(x) - ∫ [u' {∫v dx} dx] dx
⇒ ∫ logx.1.dx =  logx. ∫1.dx - ∫ ((logx)'.∫ 1.dx).dx
⇒ ∫ logx.1.dx = logx.x -∫ (1/x .x).dx
⇒ ∫ logx.1.dx = xlogx - ∫ 1.dx
⇒ ∫ logx.dx = x logx - x + C

Which is the required integration of logarithmic function.

Integration of Inverse Trigonometric Function (tan^-1 x)

Integration of Inverse Trigonometric Functions (tan^-1 x) is achieved using Integration by part formula. The integration is discussed below,

∫ tan^-1x.dx = ∫tan^-1x.1.dx

Taking tan^-1 x as the first function and 1 as the second function.

Using ∫u.v dx = u ∫ v d(x) - ∫ [u' {∫v dx} dx] dx
⇒ ∫tan^-1x.1.dx = tan^-1x.∫1.dx - ∫((tan^-1x)'.∫ 1.dx).dx
⇒ ∫tan^-1x.1.dx = tan^-1x. x - ∫(1/(1 + x²).x).dx
⇒ ∫tan^-1x.1.dx = x. tan^-1x - ∫ 2x/(2(1 + x²)).dx
⇒ ∫tan^-1x.dx = x. tan^-1x - ½.log(1 + x2) + C

Which is the required integration of Inverse Trigonometric Function.

Integration by Parts Formulas

We can derive the integration of various functions using the integration by parts concept. Some of the important formulas derived using this technique are

• ∫ e^x(f(x) + f'(x)).dx = e^xf(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

Integration By Parts Examples

Example 1: Find ∫ e^x x dx.

Solution:

Let I =  ∫ e^x x dx

Choosing u and v using ILATE rule

u = x
v = e^x

Differentiating u

u'(x) = d(u)/dx

⇒ u'(x) = d(x)/dx
⇒ u'(x) = 1

∫v dx = ∫e^x dx = e^x

Using the Integration by part formula,

⇒ I = ∫ e^x x dx 
⇒ I = x ∫e^x dx − ∫1 (∫ e^x dx) dx
⇒ I = xe^x − e^x + C
⇒ I = e^x(x − 1) + C

Example 2: Calculate ∫ x sin x dx.

Solution:

Let I = ∫ x sin x dx

Choosing u and v using ILATE rule

u = x
v = sin x

Differentiating u

u'(x) = d(u)/dx
⇒ u'(x) = d(x)/dx
⇒ u'(x) = 1

Using the Integration by part formula,

⇒ I = ∫ x sin x dx 
⇒ I = x ∫sin x dx − ∫1 ∫(sin x dx) dx
⇒ I = − x cos x − ∫−cos x dx
⇒ I = − x cos x + sin x + C

Example 3: Find ∫ sin⁻¹ x  dx.

Solution:

Let I=  ∫ sin⁻¹ x  dx

⇒ I =  ∫ 1.sin⁻¹ x  dx

Choosing u and v using ILATE rule

u = sin⁻¹ x
v = 1

Differentiating u

u'(x) = d(u)/dx
⇒ u'(x) = d(sin⁻¹ x )/dx 
⇒ u'(x) = 1/√(1 − x ²) 

Using the Integration by part formula,

⇒ I = ∫ sin⁻¹ x dx 
⇒ I = sin⁻¹ x ∫ 1 dx  − ∫ 1/√(1 − x ²)  ∫(1 dx)  dx
⇒ I  = x sin⁻¹ x  − ∫( x/√(1 − x ² ) )dx

Let, t = 1 − x ² 

Differentiating both sides

 dt = −2x dx
⇒ −dt/2 = x dx
⇒ I = ∫ sin⁻¹ x dx =  x sin⁻¹ x  − ∫−(1/2√t ) dt
⇒ I   = x sin⁻¹ x  + 1/2∫t^−1/2 dt
⇒ I  = x sin⁻¹ x + t^1/2 + C
⇒ I  = x sin⁻¹ x + √(1 − x² ) + C