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Monday, 4 August 2014

CHAPTER 2- Integration by Simple Rearrangements

The rearrangement technique is best illustrated through examples.
     Example: 1    

Evaluate \int {\dfrac{{\cos x - \cos 2x}}{{1 - \cos x}}dx} .
Solution: 1

We can try expanding \cos 2x by the half angle formula:
\int {\dfrac{{\cos x - \cos 2x}}{{1 - \cos x}}dx}  = \int {\dfrac{{\cos x - \left( {2{{\cos }^2}x - 1} \right)}}{{1 - \cos x}}dx}
 = \int {\dfrac{{\left( {2\cos x + 1} \right)\left( {1 - \cos x} \right)}}{{\left( {1 - \cos x} \right)}}dx}
 = \int {\left( {2\cos x + 1} \right)dx}
 = 2\sin x + x + C.
Observe how the rearrangement we used led to a simpler expression that was easily integrable.
     Example: 2     

Evaluate \int {\dfrac{1}{{\sin \left( {x - a} \right)\cos \left( {x - b} \right)}}dx} .
Solution: 2

The denominator is of the form \sin P\cos Q, where P and Q are variable; but notice an important fact: P - Q is a constant. This should give us the required hint:
\int {\dfrac{1}{{\sin \left( {x - a} \right)\cos \left( {x - b} \right)}}dx = \dfrac{1}{{\cos \left( {a - b} \right)}}\int {\dfrac{{\cos \left( {a - b} \right)}}{{\sin \left( {x - a} \right)\cos \left( {x - b} \right)}}} \,dx}
 = \dfrac{1}{{\cos \left( {a - b} \right)}}\int {\dfrac{{\cos \left\{ {\left( {x - b} \right) - \left( {x - a} \right)} \right\}}}{{\sin \left( {x - a} \right)\cos \left( {x - b} \right)}}dx}
 = \dfrac{1}{{\cos \left( {a - b} \right)}}\int {\dfrac{{\cos \left( {x - b} \right)\cos \left( {x - a} \right) + \sin \left( {x - b} \right)\sin \left( {x - a} \right)}}{{\sin \left( {x - a} \right)\cos \left( {x - b} \right)}}dx}
 = \dfrac{1}{{\cos \left( {a - b} \right)}}\int {\left\{ {\cot \left( {x - a} \right) + \tan \left( {x - b} \right)} \right\}dx}
 = \dfrac{1}{{\cos \left( {a - b} \right)}}\left\{ {\ln \left| {\sin \left( {x - a} \right)} \right| - \ln \left| {\cos \left( {x - b} \right)} \right|} \right\} + C
 = \dfrac{1}{{\cos \left( {a - b} \right)}}\ln \left| {\dfrac{{\sin \left( {x - a} \right)}}{{\cos \left( {x - b} \right)}}} \right| + C
What we had to do in this question was therefore to realise that since P - Qis a constant, an introduction of the term \cos (P - Q) in the numerator would lead to cancellations and simple ‘cot’ and ‘tan’ terms which can easily be integrated.
     Example: 3     

Evaluate \int {\dfrac{{{x^3}}}{{{{\left( {x + 1} \right)}^2}}}dx} .
Solution: 3

The numerator has a degree higher then the denominator which hints that some reduction of this rational expression is possible. This reduction can be accomplished if we somehow rearrange the numerator in such a way that it leads to a cancellation of common factors with the denominator; since the denominator is {(x + 1)^2}, we try to rearrange the numerator in terms of (x + 1):
\int {\dfrac{{{x^3}}}{{{{\left( {x + 1} \right)}^2}}}dx}  = \int {\dfrac{{\left( {{x^3} + 1} \right) - 1}}{{{{\left( {x + 1} \right)}^2}}}dx}
 = \int {\left\{ {\dfrac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}\,\,dx}
 = \int {\left\{ {\dfrac{{{x^2} - x + 1}}{{\left( {x + 1} \right)}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}\,\,dx}
 = \int {\left\{ {\dfrac{{{x^2} - x - 2 + 3}}{{x + 1}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}\,\,} dx
 = \int {\left\{ {\dfrac{{\left( {x + 1} \right)\left( {x - 2} \right) + 3}}{{\left( {x + 1} \right)}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}\,\,} dx
 = \int {\left\{ {\left( {x - 2} \right) + \dfrac{3}{{x + 1}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}} \,\,dx
 = \dfrac{{{x^2}}}{2} - 2x + 3\ln \left( {x + 1} \right) + \dfrac{1}{{x + 1}} + C
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