ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 4- MORE Worked Out Examples

Monday 4 August 2014

CHAPTER 4- MORE Worked Out Examples

Example: 1

Evaluate $\int\limits_{ - \infty }^{ + \infty } {\dfrac{1}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}\,\,dx}$

Solution: 1

$\dfrac{1}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}$ can be rewritten as $\dfrac{1}{{\left( {{b^2} - {a^2}} \right)}}\left\{ {\dfrac{1}{{{x^2} + {a^2}}} - \dfrac{1}{{{x^2} + {b^2}}}} \right\}$ Therefore, the given integral becomes

	$\dfrac{1}{{\left( {{b^2} - {a^2}} \right)}}\left\{ {\int\limits_{ - \infty }^{ + \infty } {\dfrac{1}{{\left( {{x^2} + {a^2}} \right)}}dx} - \int\limits_{ - \infty }^{ + \infty } {\dfrac{1}{{\left( {{x^2} + {b^2}} \right)}}dx} } \right\}$
	$= \dfrac{1}{{\left( {{b^2} - {a^2}} \right)}}\left\{ {\left. {\dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a}} \right\|_{ - \infty }^{ + \infty } - \left. {\dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}} \right\|_{ - \infty }^{ + \infty }} \right\}$
	$= \dfrac{1}{{\left( {{b^2} - {a^2}} \right)}}\left\{ {\dfrac{\pi }{a} - \dfrac{\pi }{b}} \right\}$
	$= \dfrac{\pi }{{ab\left( {a + b} \right)}}$

Sometimes, to modify an integral, an appropriate substitution has to be used; the same way we did in the unit on Indefinite Integration. For example, integrals containing the expression $({x^2} + {a^2})$ can be simplified (or modified) using the substitution $x = a\tan \theta$ .

For evaluating a definite integral too, we can use the appropriate substitution, provided we change the limits of integration accordingly also. This will become clear in subsequent examples.

Example: 2

If ${I_n} = \int\limits_0^{x/4} {{{\tan }^n}x\,dx,\,}$ prove that

	(a) ${I_n} + {I_{n - 2}} = \dfrac{1}{{n - 1}}$
	(b) $\dfrac{1}{{n + 1}}\,\,\, < \,\,\,2{I_n} < \dfrac{1}{{n - 1}}$

Solution: 2-(a)

	${I_n} + {I_{n - 2}} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} + \int\limits_0^{\pi /4} {{{\tan }^{n - 2}}x\,dx}$
	$= \int\limits_0^{\pi /4} {\left( {{{\tan }^n}x + {{\tan }^{n - 2}}x} \right)} \,dx$
	$= \int\limits_0^{\pi /4} {{{\tan }^{n - 2}}x\left( {{{\tan }^2}x + 1} \right)} \,\,dx$
	$= \int\limits_0^{\pi /4} {{{\tan }^{n - 2}}x{{\sec }^2}x} \,\,dx$

The substitution $\tan x = t$ can now be used to simplify this integral. However, we must change the limits of integration according to this substitution:

	$\tan x = t\, \,\,\, \Rightarrow \,\,\,\, {\sec ^2}x\,dx = dt\,$
	If $x = 0 \,\,\,\, \Rightarrow \,\,\,\, t = 0$
	If $\,\,x = \dfrac{\pi }{4}\, \,\,\, \Rightarrow \,\,\,\, t = 1\,$

Thus, the modified integral (in terms of the new variable $t$ ) is:

	${I_n} + {I_{n - 2}} = \int\limits_0^1 {{t^{n - 2}}\,dt}$
	$\left. { = \dfrac{{{t^{n - 1}}}}{{n - 1}}} \right\|_0^1$
	$= \dfrac{1}{{n - 1}}$

Solution: 2-(b)

In the integral that we are considering, the limits of integration are $0$ to $\dfrac{\pi }{4}$ i.e., $x \in \left[ {0,\,\,\dfrac{\pi }{4}} \right]$

In this interval, $\tan x < 1$ . Thus,

${\tan ^{n + 2}}x < {\tan ^n}x < \,{\tan ^{n - 2}}x\,\,\,\,\,\,\forall \,\,\,\,x \in \left( {0,\,\,\dfrac{\pi }{4}} \right)$

From property ( $4$ ), we can therefore say that:

	$\int\limits_0^{\pi /4} {{{\tan }^{n + 2}}x\,dx < } \int\limits_0^{\pi /4} {{{\tan }^n}x < } \int\limits_0^{\pi /4} {{{\tan }^{n - 2}}x}$
	or ${I_{n + 2}}\,\, < \,\,{I_n} < {I_{n - 2}}$
	$\Rightarrow \,\,\,\, {I_n}\,\, + \,\,{I_{n + 2}}\,\, < 2{I_n}\,\,\, < {I_n} + {I_{n - 2}}\,$	$\ldots(1)$

Using the result of part ( $a$ ) for the first and third terms in ( $1$ ), we get our desired result:

$\dfrac{1}{{n + 1}}\,\, < \,\,2{I_n}\,\,\, < \,\,\,\dfrac{1}{{n - 1}}$

Example: 3

For $x > 0$ , let $f\left( x \right) = \int\limits_1^x {\dfrac{{\ln t}}{{1 + t}}\,\,dt.}$ . Find the function $f\left( x \right) + f\left( {\dfrac{1}{x}} \right)$ and show that $f\left( e \right) + f\left( {\dfrac{1}{e}} \right) = \dfrac{1}{2}.$

Solution: 3

Observe carefully the form of the function $f(x)$ : It is in the form of an integral (of another function), with the lower limit being fixed and the upper limit being the variable $x$ . As $x$ varies, $f(x)$ will correspondingly vary.

One approach that you might contemplate to solve this question is evaluate the anti-derivative $g(t)$ of $\dfrac{{\ln t}}{{1 + t}}$ and then evaluate $g\left( x \right) - g\left( 1 \right)$ which will become $f(x)$ . However, this will become unnecessarily cumbersome (Try it!). We can, instead, proceed as follows:

	$f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = \int\limits_1^x {\dfrac{{\ln t}}{{1 + t}}dt} \,\, + \int\limits_1^{1/x} {\dfrac{{\ln t}}{{1 + t}}dt}$
	$= {I_1} + {I_2}$

Notice that the limits of integration of ${I_1}$ and ${I_2}$ are different. If they were the same, we could have added ${I_1}$ and ${I_2}$ easily. So we try to make them the same: in ${I_2}$ , if we let and $t$ varies from $1$ to $y$ will vary from $1$ to $x$ . This substitution will therefore make the limits of integration of ${I_2}$ the same as those of ${I_1}$ :

	$t = \dfrac{1}{y}$
	$dt = - \dfrac{1}{{{y^2}}}dy$
	$t = 1 \,\,\,\, \Rightarrow \,\,\,\, y = 1$
	$t = \dfrac{1}{x}\, \,\,\, \Rightarrow \,\,\,\, y = x$
	${I_2} = \int\limits_1^{1/x} {\dfrac{{\ln t}}{{1 + t}}\,\,dt}$
	$= - \int\limits_1^x {\dfrac{{\ln \left( {1/y} \right)}}{{1 + \left( {1/y} \right)}}\,\,\dfrac{1}{{{y^2}}}dy}$
	$= \int\limits_1^x {\dfrac{{\ln y}}{{y\left( {1 + y} \right)}}dy}$

${I_1}$ and ${I_2}$ can now be easily added:

	$\begin{array}{l} = {I_1} + {I_2} = \int\limits_1^x {\left\{ {\dfrac{{\ln t}}{{1 + t}} + \dfrac{{\ln t}}{{t\left( {1 + t} \right)}}} \right\}} \,\,dt\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{array}$
	$= \int\limits_1^x {\dfrac{{\left( {{\mathop{\rm l}\nolimits} + t} \right)\ln t}}{{t\left( {1 + t} \right)}}} \,\,dt$
	We used $t$ instead of $y$ in ${I_2}$ . This doesn’t make a difference; $y$ is the variable of integration; it can be replaced with any other variable as long as the limits of integration are the same
	$= \int\limits_1^x {\dfrac{{\ln t}}{t}} \,\,dt$

The final expression shows how simplified ${I_1} + {I_2}$ has become. We let $\ln t = z\,\, \Rightarrow \,\,\,\dfrac{1}{t}\,\,dt = dz$ and the limits of integration become $0$ to $\ln x$ .