ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 3- Worked Out Examples

Monday, 4 August 2014

CHAPTER 3- Worked Out Examples

Example: 1

Evaluate $\int {\dfrac{x}{{\sqrt {x + a} + \sqrt {x + b} }}dx}$ .

Solution: 1

The form of the expression in the denominator clearly hints that a reduction is possible by rationalization which would lead to a constant term in the denominator:

	$\int {\dfrac{x}{{\sqrt {x + a} + \sqrt {x + b} }}dx} = \int {\dfrac{{x\left\{ {\sqrt {x + a} - \sqrt {x + b} } \right\}}}{{\left( {x + a} \right) - \left( {x + b} \right)}}} dx$
	$= \dfrac{1}{{\left( {a - b} \right)}}\int {\left\{ {x\sqrt {x + a} - x\sqrt {x + b} } \right\}dx}$
	$= \dfrac{1}{{a - b}}\int {\left\{ {\left( {x + a - a} \right)\sqrt {x + a} - \left( {x + b - b} \right)\sqrt {x + b} } \right\}dx}$
	$= \dfrac{1}{{a - b}}\int {\left\{ {{{\left( {x + a} \right)}^{3/2}} - {{\left( {x + b} \right)}^{3/2}} - a{{\left( {x + a} \right)}^{1/2}} + b{{\left( {x + b} \right)}^{1/2}}} \right\}dx}$
	$= \dfrac{1}{{a - b}}\left\{ {\dfrac{{{{\left( {x + a} \right)}^{5/2}}}}{{5/2}} - \dfrac{{{{\left( {x + b} \right)}^{5/2}}}}{{5/2}} - \dfrac{{a{{\left( {x + a} \right)}^{3/2}}}}{{3/2}} + \dfrac{{b{{\left( {x + b} \right)}^{3/2}}}}{{3/2}}} \right\} + C$

The first simplification by rationalization led to an expression which involved two terms of the form $x\sqrt {x + a}$ ; to integrate these terms, we wrote the $x$ outside the root as $(x + a - a)$ so that a final expression is obtained which contains only terms of the form ${\left( {x + k} \right)^n}$ ; these could then be integrated easily.

Example: 2

Evaluate $\int {\dfrac{{1 + \cos 4x}}{{\cot x - \tan x}}dx}$

Solution: 2

We simply both the numerator and the denominator separately :

	$\int {\dfrac{{1 + \cos 4x}}{{\cot x - \tan x}}dx} = \int {\dfrac{{2{{\cos }^2}2x}}{{\dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}}}}dx}$
	$= \int {\dfrac{{2\sin x\cos x{{\cos }^2}2x}}{{{{\cos }^2}x - {{\sin }^2}x}}dx}$
	$= \int {\dfrac{{\sin 2x \cdot {{\cos }^2}2x}}{{\cos 2x}}dx}$
	$= \int {\sin 2x\cos 2x} \,\,\,\,dx$
	$= \dfrac{1}{2}\int {\sin 4x\,\,dx}$
	$= - \dfrac{1}{8}\cos 4x + C$

Example: 3

Evaluate $\int {\tan x\tan 2x\tan 3x} \,dx$ .

Solution: 3

Notice that $3x = 2x + x$ , so that

	$\tan 3x = \tan \left( {2x + x} \right)$
	$= \dfrac{{\tan 2x + \tan x}}{{1 - \tan 2x\tan x}}$
	$\Rightarrow \,\,\,\tan x\tan 2x\tan 3x = \tan 3x - \tan 2x - \tan x$

The required integral is now easy to evaluate :

	$\int {\tan x\tan 2x\tan 3x} \,\,dx = \int {\left\{ {\tan 3x - \tan 2x - \tan x} \right\}} \,\,dx$
	$= \dfrac{1}{3}\ln \left\| {\sec 3x} \right\| - \dfrac{1}{2}\ln \left\| {\sec 2x} \right\| - \ln \left\| {\sec x} \right\| + C$

Example: 4

Evaluate $\int {\dfrac{{\sin \left( {x - a} \right)}}{{\sin \left( {x - b} \right)}}dx}$ .

Solution: 4

Taking cue from Example- $2$ , our aim should be to somehow get rid of the variable term $\sin (x - b)$ in the denominator; to do this, we write the numerator $\sin \left( {x - a} \right)$ as $\sin \left\{ {\left( {x - b} \right) - \left( {a - b} \right)} \right\}$ :

	$\int {\dfrac{{\sin \left( {x - a} \right)}}{{\sin \left( {x - b} \right)}}dx = } \int {\dfrac{{\sin \left\{ {\left( {x - b} \right) - \left( {a - b} \right)} \right\}}}{{\sin \left( {x - b} \right)}}dx}$
	$= \int {\dfrac{{\sin \left( {x - b} \right)\cos \left( {a - b} \right) - \cos \left( {x - b} \right)\sin \left( {a - b} \right)}}{{\sin \left( {x - b} \right)}}dx}$
	$= \int {\left\{ {\cos \left( {a - b} \right) - \sin \left( {a - b} \right)\cot \left( {x - b} \right)} \right\}dx}$
	$= x\cos \left( {a - b} \right) - \sin \left( {a - b} \right)\ln \left\| {\sin \left( {x - b} \right)} \right\| + C$

Monday, 4 August 2014

CHAPTER 3- Worked Out Examples

No comments:

https://www.youtube.com/TarunGehlot