# Definite Integration Set 2

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Example -3

If $$m,n \in \mathbb{N},$$ prove that $${I_{m,n}} = \int\limits_0^1 {{x^m}{{(1 - x)}^n}dx = \frac{{m!n!}}{{(m + n + 1)!}}}$$

Solution: It is clear from the final result which we need to obtain that we require a recursive relation involving $${I_{m,n}}$$ and a lower order integral (w.r.t one of the two variables m or n).

\begin{align}&{I_{m,n}} = \int\limits_0^1 {\mathop {{{(1 - x)}^n}}\limits_{\uparrow \atop{\scriptstyle{\rm{Ist}}\atop\scriptstyle{\rm{Function}}}} \mathop {{x^m}}\limits_{\scriptstyle{\rm{IInd}}\atop\scriptstyle{\rm{Function}}} dx} \\\,\,\,\,\,\,\,\,\, &\quad\;\;= \left. {\frac{{{{(1 - x)}^n}{x^{m + 1}}}}{{m + 1}}} \right|_0^1 + \frac{n}{{m + 1}}\int\limits_0^1 {{{(1 - x)}^{n - 1}}{x^{m + 1}}dx} \\\,\,\,\,\,\,\,\,\, &\quad\;\;= 0 + \frac{n}{{m + 1}}{I_{m + 1,n - 1}}\end{align}

Thus, the required recursive relation is

${I_{m,n}} = \frac{n}{{m + 1}}{I_{m + 1,n - 1}}$

We use this relation repeatedly now till n reduces to 0:

\begin{align}&{I_{m,n}} = \frac{n}{{m + 1}}{I_{m + 1,n - 1}}\\\,\,\,\,\,\,\,\,\, &\;\;\quad= \frac{{n(n - 1)}}{{(m + 1)(m + 2)}}{I_{m + 2,n - 2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\;\;\qquad\;\;\qquad\vdots \\\,\,\,\,\,\,\,\,\, &\;\;\quad= \frac{{n(n - 1).............1}}{{(m + 1)(m + 2)....(m + n)}}{I_{m + n,0}}\\\,\,\,\,\,\,\,\,\, &\;\;\quad= \frac{{m!n!}}{{(m + n)!}}{I_{m + n,0}}\end{align}

$${I_{m + n,0}}$$ is easy to evaluate. Verify that it is $$\frac{1}{{m + n + 1}}$$

Thus, \begin{align}{I_{m,n}} = \frac{{m!n!}}{{(m + n + 1)!}}\end{align}

Example -4

Evaluate $$I = \int\limits_0^1 {{{\cot }^{ - 1}}(1 + {x^2} - x)dx}$$

Solution: Before reading the solution below, you are advised to try simplifying I directly using the properties and techniques developed in this chapter. You’ll realise the difficulty of the task.

In this example, instead of direct application of any property, a non-trivial manipulation is first required to simplify I before applying any properties. That manipulation is now described:

\begin{align}& I = \int\limits_0^1 {{{\cot }^{ - 1}}(1 + {x^2} - x)dx} \\ \,\,\,\, &\;= \int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{1}{{1 + x(x - 1)}}} \right)dx} \\ \,\,\,\, &\;= \int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{{x - (x - 1)}}{{1 + x(x - 1)}}} \right)dx\qquad\qquad \left\{ \begin{gathered} {\text{We wrote the numerator '1'}} \\ {\text{as '}}x - (x - 1)' \\ \end{gathered} \right\}} \\\,\,\,\, &\;= \int\limits_0^1 {\left\{ {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}(x - 1)} \right\}dx \qquad\qquad \left\{ {\because \,\,\,{{\tan }^{ - 1}}\left( {\frac{{A - B}}{{1 + AB}}} \right) = {{\tan }^{ - 1}}A - {{\tan }^{ - 1}}B} \right\}} \\ \,\,\,\, &\;= \int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx - \int\limits_0^1 {{{\tan }^{ - 1}}(x - 1)dx} } \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dots(1) \\ \end{align}

Observe how simple the integral I has now become. A further simplification is possible by the application of property - 9 on the second integral in (1):

\begin{align}&\int\limits_0^1 {{{\tan }^{ - 1}}(x - 1) = \int\limits_0^1 {{{\tan }^{ - 1}}\left( {(1 - x) - 1} \right)dx} } \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\;\;= - \int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx} \end{align}

Thus, $$I = 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx}$$

We can now proceed to evaluate I using integration by parts:

\begin{align}&I = 2\left\{ {\left. {x{{\tan }^{ - 1}}x} \right|_0^1 - \int\limits_0^1 {\frac{x}{{1 + {x^2}}}dx} } \right\}\\\,\,\,\, &\;= 2\left\{ {\frac{\pi }{4} - \frac{1}{2}\int\limits_1^2 {\frac{{dt}}{t}} } \right\} & & & \left( \begin{array}{l}{\rm{We\;used\;the\;substitution}}\\1 + {x^2} = t\end{array} \right)\\\,\,\,\, &\;= \left. {\frac{\pi }{2} - (\ln t)} \right|_1^2\\\,\,\,\, &\;= \frac{\pi }{2} - \ln 2\end{align}

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