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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 $P - Q$is 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 $(x + 1)$:
 $\int {\dfrac{{{x^3}}}{{{{\left( {x + 1} \right)}^2}}}dx} = \int {\dfrac{{\left( {{x^3} + 1} \right) - 1}}{{{{\left( {x + 1} \right)}^2}}}dx}$ $= \int {\left\{ {\dfrac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}\,\,dx}$ $= \int {\left\{ {\dfrac{{{x^2} - x + 1}}{{\left( {x + 1} \right)}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}\,\,dx}$ $= \int {\left\{ {\dfrac{{{x^2} - x - 2 + 3}}{{x + 1}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}\,\,} dx$ $= \int {\left\{ {\dfrac{{\left( {x + 1} \right)\left( {x - 2} \right) + 3}}{{\left( {x + 1} \right)}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}\,\,} dx$ $= \int {\left\{ {\left( {x - 2} \right) + \dfrac{3}{{x + 1}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right\}} \,\,dx$ $= \dfrac{{{x^2}}}{2} - 2x + 3\ln \left( {x + 1} \right) + \dfrac{1}{{x + 1}} + C$