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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 xoutside 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
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