Integration by Parts

Integration by Parts

When the integrand can not be reduced to standard form by substitution, we apply integration by parts. It corresponds to product rule in differentiation.

If u and v are functions of x, then

Integrating, we get

This is called the formula for integration by parts. The success of the method depends on the proper choice of u as that function which comes first in the word 'ILATE' where

I - inverse circular function

L - logarithmic function 

A - algebraic function

T - trigonometric function 

E - exponential function

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Example 2 

Evaluate

Solution

Example 3 

Evaluate

Solution 

Example 4 

Evaluate

Solution

Example 5 

Evaluate

Solution

Example 6 

Find

Solution

Example 7 

Evaluate

Solution

Example 8 

Evaluate

Solution

Example 9 

Evaluate

Solution

Put t = sin x.

Example 10 

Evaluate

Solution

Example 11 

Evaluate

Solution

Example 12 

Evaluate

Solution

Example 13 

Using integration by parts, evaluate

Solution

Repeated application of integration by parts is called Bernoulli formula.

1. Bernoulli's Formula

If u and v are differentiable functions of x, then

where primes denote differentiation and suffixes denote integration.

That is

and

If u is a polynomial in x, then Bernoulli's formula terminates.

2. Special Integrals

Solution 

It is a product of two functions.

So, using integration by parts, we evaluate the integral.

Solution 

It is a product of two functions. So, we use integration by parts to evaluate the integral. 

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Example 2 

Evaluate

Solution

Example 3 

Evaluate

Solution

Example 4 

Evaluate

Solution

Example 5 

Evaluate

Example 6 

Evaluate

Solution

Example 7 

Evaluate

Solution

3. Reduction Formula

Integrals of type cannot be evaluated directively. Applying integration by parts, we can reduce an integral with index n > 0, called the order of the integral, to an integral of the reduced order with a smaller index. The relation between the given integral and the reduced integral of lower order is called the reduction formula.

We derive the reduction formula for some standard integrals

WORKED EXAMPLES

Example 1 

Solution

Example 2 

Solution

Deduction:

If n is a non-negative integer, then prove that

Solution

Case 1: If n is even, then

Case 2: if n is odd, then

Similarly, we get

Example 3 

Solution

Example 4 

Solution

Example 5 

Show that

Solution

Example 6 

Solution

Here n = 5 is odd

Example 7 

Evaluate

Solution

Example 8 

Evaluate

Solution

Example 9 

Solution

Here, n = 7 is odd

EXERCISE 

Evaluate following integrals by integration by parts formula

ANSWERS TO EXERCISE 

4. Properties of Definite Integrals

If f(x) is a continuous and integrable function of x in [a, b], then the following properties are satisfied.

Note

That is the value of a definite integral is unaffected by the change of dummi variable, if the limits and function are the same.

Substituting in (1), we get

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Example 2

Solution

Example 3 

If ƒ is continuous on [0, π] show that

Solution

Example 4 

Show that

Solution

Note Since the right hand side is independent of n, this is true for all n.

Example 5 

Prove that

Solution

Example 6 

Prove that

Solution

Example 7 

Evaluate

Solution

Example 8 

Prove that

Solution

Note In the interval (0, π) we cannot put t = tan x as it is not increasing there and discontinuous at x = π/2. So, we reduced the interval from (0, π) to [0, π/2] by property 6, so that in (0, π/2), tan x is strictly increasing.

Example 9 

Evaluate

Solution

Example 10 

Evaluate

Solution

Here the interval is [0,4],

Example 11 

Evaluate

Solution

Example 12 

Evaluate

Solution

Example 13 

Evaluate

Solution

We know

5. Leibnitz Rule for Derivative of Integral

We shall now find derivatives of integrals using part-I of Fundamental theorem of calculus and we obtain Leibnitz rule for derivative.

WORKED EXAMPLES

Example 1 

If g(x) =

Solution 

Example 2 

Solution

Example 3 

Solution

Example 4 

Solution

Example 5 

Solution

Example 6 

Solution

Example 7 

Solution

Example 8 

Solution

Example 9 

Solution

Example 10 

The function f is continuous and satisfies

Solution 

Differentiate with respect to x, we get

To find c, substitute for f in the given equation f(t) = 2t15

EXERCISE

Evaluate the following integrals

ANSWERS TO EXERCISE 

6. Definite integral as a limit of a sum

Working Rule

WORKED EXAMPLES

Example 1 

Show that

Solution 

Example 2 

Solution 

Example 3 

Show that

Solution 

Let

To convert the product into sum, take logarithm on both sides

Example 4 

Solution 

Integrating by parts, we get

EXERCISE

Evaluate the following limits as integrals

ANSWERS TO EXERCISE