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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.
{I_1} + {I_2} = \int\limits_0^{\ln x} {z\,dz}
 = \dfrac{1}{2}{\left( {\ln x} \right)^2}
Thus,
f\left( e \right) + f\left( {\dfrac{1}{e}} \right) =\dfrac{1}{2}{\left( \ln e \right)^2}
 = \dfrac{1}{2}
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