ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 4- TRIGONOMETRIC INTEGRALS

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)
	$= \dfrac{3}{{32}}\sin 2x - \dfrac{1}{{32}}\sin 6x$