Basic Examples on Definite Integrals Set 3

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Example –6

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

Solution: 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 \(\frac{{\ln t}}{{1 + t}}\) 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:

\[\begin{align}&f\left( x \right) + f\left( {\frac{1}{x}} \right) = \int\limits_1^x {\frac{{\ln t}}{{1 + t}}dt} \,\, + \int\limits_1^{1/x} {\frac{{\ln t}}{{1 + t}}dt} \\ &\qquad\qquad\qquad\;\;= {I_1} + {I_2}\end{align}\]

Notice that the limits of integration of I 1 and I2 are different. If they were the same, we could have added I1 and I2 easily. So we try to make them the same: in  I2 , if we let \(t = \frac{1}{y},\) and t varies from 1 to \(\frac{1}{x},\) y will vary from 1 to x. This substitution will therefore make the limits of integration of I2 the same as those of I1 :

\[\begin{align}&t = \frac{1}{y}\\&dt = - \frac{1}{{{y^2}}}dy\\&t = 1 \quad\;\;\Rightarrow \quad y = 1\\&t = \frac{1}{x}  \quad \Rightarrow \quad y = x\\&\qquad\qquad \quad{I_2} = \int\limits_1^{1/x} {\frac{{\ln t}}{{1 + t}}\,\,dt} \\& \qquad\qquad= - \int\limits_1^x {\frac{{\ln \left( {1/y} \right)}}{{1+ \left( {1/y} \right)}}\,\,\frac{1}{{{y^2}}}dy} \\&\qquad\qquad= \int\limits_1^x {\frac{{\ln y}}{{y\left( {1 + y} \right)}}dy}\end{align}\]

\(I_1\) and \(I_2\) can now be easily added:

\[\begin{array}{l} = {I_1} + {I_2} = \int\limits_1^x {\left\{ {\frac{{\ln t}}{{1 + t}} + \frac{{\ln t}}{{t\left( {1 + t} \right)}}} \right\}} \,\,dt\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_1^x {\frac{{\left( {{\mathop{\rm l}\nolimits} + t} \right)\ln t}}{{t\left( {1 + t} \right)}}} \,\,dt\,\left\{ \begin{array}{l}{\rm{We\;used\;}}t{\rm{\;instead\;of\;}}y{\rm{ \;in\;}}{\;I_{\rm{\;2}}}{\rm{. }}\\{\rm{\;This\;doesn't\;make\;a\;difference;\;}}\\ y{\rm{\;is\;the\;variable\;of\;integration;\;}}\\{\rm{it\;can\;be\;replaced\;with\;any\; other}}\\{\rm{\;variable\;as\;long\;as\;the\;limits\;of\;}}\\{\rm{integration\;are\;the \;same}}{\rm{. }}\end{array} \right\}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_1^x {\frac{{\ln t}}{t}} \,\,dt\,\end{array}\]

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

\[\begin{align}&{I_1} + {I_2} = \int\limits_0^{\ln x} {z\,dz} \\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad= \frac{1}{2}{\left( {\ln x} \right)^2}\end{align}\]

Thus,

\[\begin{align}&f\left( e \right) + f\left( {\frac{1}{e}} \right) = {\left( {\frac{{\ln e}}{2}} \right)^2}\\\, &\qquad\qquad\qquad\;\;= \frac{1}{2}\end{align}\]

Example –7

Evaluate \(\int\limits_0^{\pi /2} {{{\left( {\sqrt {\sin x} \, + \sqrt {\cos x} \,} \right)}^{ - 4}}\,\,dx} \)

Solution: The given integral can be modified into an (easily) integrable form by expressing it in a form involving tan x and sec x.

\[\begin{align}&I = \int\limits_0^{\pi /2} {\frac{1}{{{{\left( {\sqrt {\sin x} + \sqrt {\cos x} } \right)}^4}}}} \,\,dx\\\,\,\, &= \int\limits_0^{\pi /2} {\frac{1}{{{{\cos }^2}x{{\left( {1 + \sqrt {\tan x} \,} \right)}^4}}}} \,\,dx\\\,\,\, &= \int\limits_0^{\pi /2} {\frac{{{{\sec }^2}x}}{{{{\left( {1 + \sqrt {\tan x} \,} \right)}^4}}}} \,\,dx\end{align}\]

The substitution tan \(x=y^2\) can now be used.

\[\begin{align}\Rightarrow \quad {\sec ^2}x\,dx\,\, = \,2ydy\\x = 0 \quad\;\; \Rightarrow\;\; \quad y = 0\\x = \pi /2\, \quad \Rightarrow \quad y = \infty\end{align}\]

\[\begin{align}&I = 2\int\limits_0^\infty {\frac{{y\,dy}}{{{{\left( {1 + y} \right)}^4}}}} \\ &\;\;= 2\int\limits_0^\infty {\frac{{\left( {1 + y} \right) - 1}}{{{{\left( {1 + y} \right)}^4}}}} \,\,dy\\&\;\;= 2\int\limits_0^\infty {\left\{ {\frac{1}{{{{\left( {1 + y} \right)}^3}}} - \frac{1}{{{{\left( {1 + y} \right)}^4}}}} \right\}} \,\,dy\\&\;\;= 2\left\{ {\left. {\frac{{{{\left( {1 + y} \right)}^{ - 2}}}}{{ - 2}}} \right|_0^\infty - \left. {\frac{{{{\left( {1 + y} \right)}^{ - 3}}}}{{ - 3}}} \right|_0^\infty } \right\}\,\\ &\;\;= 2\left( {\frac{1}{2} - \frac{1}{3}} \right)\\ &\;\;= \frac{1}{3}\end{align}\]