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Monday, 4 August 2014

CHAPTER 4- TRIGONOMETRIC INTEGRALS

 Example: 1     

Evaluate the following integrals:
(a) \int {{{\cos }^3}x\,\,dx}
(b) \int {{{\cos }^4}x\,\,dx}
(c) \int {\sin 2x\cos 4x\cos 5x\,\,dx}
(d) \int {{{\sin }^3}x{{\cos }^3}x\,\,dx}
Solution: 1-(a)

We know the integral of \cos x; we must express the cubic cos term ({\cos ^3}x)in terms of linear cos terms; this can be done using the triple angle formula :
\cos 3x = 4{\cos ^3}x - 3\cos x
 \Rightarrow  \,\,\,\, {\cos ^3}x = \dfrac{1}{4}\left\{ {3\cos x + \cos 3x} \right\}
 \Rightarrow  \,\,\,\, \int {{{\cos }^3}x\,\,dx = \dfrac{3}{4}\sin x + \dfrac{1}{{12}}\sin 3x + C}
Solution: 1-(b)

Here again, we need to express the fourth degree cos term \left( {{{\cos }^4}x} \right) in terms of linear cos terms:
{\cos ^4}x = {\left( {{{\cos }^2}x} \right)^2}
 = {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)^2}
 = \dfrac{1}{4} + \dfrac{{{{\cos }^2}2x}}{4} + \dfrac{1}{2}\cos 2x
 = \dfrac{1}{4} + \dfrac{{\left( {1 + \cos 4x} \right)}}{8} + \dfrac{1}{2}\cos 2x
 = \dfrac{3}{8} + \dfrac{1}{2}\cos 2x + \dfrac{1}{8}\cos 4x
 \Rightarrow  \,\,\,\, \int {{{\cos }^4}x\,\,dx = \dfrac{{3x}}{8} + \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}} + C}
Solution: 1-(c)

\sin 2x\cos 4x\cos 5x = \dfrac{1}{2}\left( {2\sin 2x\cos 4x} \right)\cos 5x
 = \dfrac{1}{2}\left( {\sin 6x - \sin 2x} \right)\cos 5x
 = \dfrac{1}{4}\left\{ {\left( {2\sin 6x\cos 5x} \right) - \left( {2\sin 2x\cos 5x} \right)} \right\}
 = \dfrac{1}{4}\left\{ {\sin 11x + \sin x - \sin 7x + \sin 3x} \right\}
 = \dfrac{1}{4}\left\{ {\sin x + \sin 3x - \sin 7x + \sin 11x} \right\}
 \Rightarrow  \,\,\, \,\int {\sin 2x\cos 4x\cos 5x\,\,dx = \dfrac{{ - \cos x}}{4} - \dfrac{{\cos 3x}}{{12}} + \dfrac{{\cos 7x}}{{28}} - \dfrac{{\cos 11x}}{{44}} + C}
Solution: 1-(d)

{\sin ^3}x{\cos ^3}x = {\left( {\sin x\cos x} \right)^3}
 = \dfrac{{{{\left( {2\sin x\cos x} \right)}^3}}}{8}
 = \dfrac{{{{\sin }^3}2x}}{8}
 = \dfrac{1}{8}\left\{ {\dfrac{{3\sin 2x - \sin 6x}}{4}} \right\}(Triple angle formula)
 = \dfrac{3}{{32}}\sin 2x - \dfrac{1}{{32}}\sin 6x
 \Rightarrow  \,\,\,\, \int {{{\sin }^3}x{{\cos }^3}x\,\,dx = \dfrac{{ - 3}}{{64}}\cos 2x + \dfrac{1}{{192}}\cos 6x + C}
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