Basic Examples on Definite Integrals Set 2

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Example - 4

Evaluate \(\begin{align}\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}\,\,dx}\end{align} \)

Solution: \(\begin{align}\frac{1}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}\end{align}\) can be rewritten as \(\begin{align}\frac{1}{{\left( {{b^2} - {a^2}} \right)}}\left\{ {\frac{1}{{{x^2} + {a^2}}} - \frac{1}{{{x^2} + {b^2}}}} \right\}\end{align}\)

Therefore, the given integral becomes

\[\begin{align}&\frac{1}{{\left( {{b^2} - {a^2}} \right)}}\left\{ {\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{\left( {{x^2} + {a^2}} \right)}}dx} - \int\limits_{ - \infty }^{ + \infty } {\frac{1}{{\left( {{x^2} + {b^2}} \right)}}dx} } \right\}\\ &= \frac{1}{{\left( {{b^2} - {a^2}} \right)}}\left\{ {\left. {\frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a}} \right|_{ - \infty }^{ + \infty } - \left. {\frac{1}{b}{{\tan }^{ - 1}}\frac{x}{b}} \right|_{ - \infty }^{ + \infty }} \right\}\\ &= \frac{1}{{\left( {{b^2} - {a^2}} \right)}}\left\{ {\frac{\pi }{a} - \frac{\pi }{b}} \right\}\\ &= \frac{\pi }{{ab\left( {a + b} \right)}} \end{align}\]

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 –5

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

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

Solution: (a)

\[\begin{align}&{I_n} + {I_{n - 2}} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} + \int\limits_0^{\pi /4} {{{\tan }^{n - 2}}x\,dx}\\&\qquad\qquad= \int\limits_0^{\pi /4} {\left( {{{\tan }^n}x + {{\tan }^{n - 2}}x} \right)} \,dx\\&\qquad\qquad= \int\limits_0^{\pi /4} {{{\tan }^{n - 2}}x\left( {{{\tan }^2}x + 1} \right)} \,\,dx\\&\qquad\qquad = \int\limits_0^{\pi /4} {{{\tan }^{n - 2}}x\left( {{{\tan }^2}x + 1} \right)} \,\,dx \end{align}\]

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:

\[\begin{align}&\tan x = t\, \quad \Rightarrow \quad {\sec ^2}x\,dx = dt\,\\&{\rm{If}}\,\,\,\,x = 0 \quad\; \Rightarrow \quad t = 0\\& {\rm{If}}\,\,\,\,x = \frac{\pi }{4}\, \quad \Rightarrow \quad t = 1\, \end{align}\]

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

\[\begin{align}&{I_n} + {I_{n - 2}} = \int\limits_0^1 {{t^{n - 2}}\,dt} \\&\quad\;\;\left. {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= \frac{{{t^{n - 1}}}}{{n - 1}}} \right|_0^1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;\;\;\;\;\;\;= \frac{1}{{n - 1}}\end{align}\]

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

In this interval, tan x < 1. Thus,

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

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

\[\begin{align}&\qquad\;\;\;{I_{n + 2}}\,\, < \,\,{I_n} < {I_{n - 2}}\\ &\Rightarrow\quad {I_n}\,\, + \,\,{I_{n + 2}}\,\, < 2{I_n}\,\,\, < {I_n} + {I_{n - 2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\end{align}\]

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

\[\frac{1}{{n + 1}}\,\, < \,\,2{I_n}\,\,\, < \,\,\,\frac{1}{{n - 1}}\]