Integration by parts method is generally used to find the integral when the integrand is a product of two different types of functions or a single logarithmic function or a single inverse trigonometric function or a function which is not integrable directly.

Integration by parts

Integration by parts method is generally used to find the integral when the integrand is a product of two different types of functions or a single logarithmic function or a single inverse trigonometric function or a function which is not integrable directly. From the formula for derivative of product of two functions we obtain this useful method of integration.

If u and v are two differentiable functions then we have

d (uv ) = vdu+udv

udv = d (uv ) - vdu

Integrating

∫udv = ∫d (uv ) - ∫vdu

∫udv = uv - ∫vdu

∫udv in terms of another integral ∫vdu and does not give a final expression for the integral ∫udv. It only partially solves the problem of integrating the product u dv . Hence the term ‘Partial Integration’ has been used in many European countries. The term “Integration by Parts” is used in many other countries as well as in our own.

The success of this method depends on the proper choice of u

(i) If integrand contains any non integrable functions directly from the formula, like logx, tan−1 x etc., we have to take these non integrable functions as u and other as dv.

(ii) If the integrand contains both the integrable function, and one of these is xn (where n is a positive integer) then take u = xn.

(iii) For other cases the choice of u is ours.

Example 11.33

Evaluate the following integrals

(i) ∫xexdx

(ii) ∫ x cos x dx

(iii) ∫log x dx

(iv) ∫sin−1x dx

Solution

(i) Let I = ∫ xe x dx.

Since x is an algebraic function and ex is an exponential function,

so take u = x then du = dx

dv = e x dx ⇒ v = ex

Applying Integration by parts, we get

∫udv = uv − ∫vdu

⇒ ∫ xex dx = xe x − ∫e x dx

That is, ∫ xe x dx = xe x − e x + c.

(ii) Let I = ∫ x cos x dx

Since x is an algebraic function and cos x is a trigonometric function,

so take u = x then du = dx

dv = cos xdx ⇒ v = sin x

Applying Integration by parts, we get

∫udv = uv − ∫vdu

⇒ ∫ x cos xdx = x sin x − ∫sin x dx

⇒ ∫ x cos xdx = x sin x + cos x + c

(iii) Let I = ∫log x dx

Take u = log x then du = 1/x dx

dv = dx ⇒ v = x

Applying Integration by parts, we get

∫udv = uv − ∫vdu

⇒ log x dx = x log x − ∫x (1/x) dx

⇒ ∫log x dx = x log x − x + c

(iv) Let I = ∫sin−1x dx

u = sin−1 (x), dv = dx

Then du = 1/ 1− x2 , v = x

Tags : Equation, Solved Example Problems | Mathematics , 11th Mathematics : UNIT 11 : Integral Calculus

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

11th Mathematics : UNIT 11 : Integral Calculus : Integration by parts | Equation, Solved Example Problems | Mathematics

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.