Introduction To Modifying Argument of Integrand Using Limits

Go back to  'Definite Integration'

In section -1, we dealt with some basic properties that definite integrals satisfy. This section continues with the development of some more properties that are not so trivial, and, when properly applied, can be extremely powerful. However, even the justifications for these properties stem from obvious physical (graphical) interpretations of functions.

(9)       \(\int\limits_0^a {f(x)\,\,dx = \int\limits_0^a {f(a - x)dx} } \)

This property says that when integrating from 0 to a, we will get the same result whether we use the function \(f(x)\) or \(f(a - x)\). The justification for this property will become clear from the figures below:

There are two ways to look at the justification of this property, as described in the figures on the left and right respectively.

Let us see how to apply this property usefully:

Example –13

Evaluate \(\begin{align}I = \int\limits_0^{\pi /2} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx}\end{align} \) .

Solution: Observe that evaluating the indefinite integral of the function above would be very tedious. Using Property-9, this integral can be immediately simplified; we use the substitution \(\begin{align} \;\;x \to \frac{\pi }{2} - x\end{align} \).

Therefore, I becomes:

\[I = \int\limits_0^{\pi /2} {\frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx} \]

Now, we add the original expression for I and this newly obtained modified expression:

\[\begin{align}2I &= \int\limits_0^{\pi /2} {\frac{{\sqrt {\sin x} + \sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx} \\&= \int\limits_0^{\pi /2} {dx} \qquad \qquad \left( \text{Quite simple now!} \right)\\&= \frac{\pi }{2}\\ \Rightarrow \qquad I &= \frac{\pi }{4}\end{align}\]

Example –14

Evaluate \(I = \int\limits_0^{\pi /4} {\ln (1 + \tan x)\,dx} \)

Solution: We again use property -9 and try to simplify this integral; use the substitution \(x \to \frac{\pi }{4} - x\) in the function to be integrated:

\[\begin{align}&I = \int\limits_0^{\pi /4} {\ln \left( {1 + \tan \left( {\frac{\pi }{4} - x} \right)} \right)\,dx} \\\,\,\, &\;\;= \int\limits_0^{\pi /4} {\ln \left( {1 + \frac{{1 - \tan x}}{{1 + \tan x}}} \right)\,dx} \\\,\,\, &\;\;= \int\limits_0^{\pi /4} {\ln \left( {\frac{2}{{1 + \tan x}}} \right)\,dx} \\\,\,\, &\;\;= \int\limits_0^{\pi /4} {\left\{ {\ln 2 - \ln (1 + \tan x)} \right\}\,\,dx} \\\,\, &\;\;= \int\limits_0^{\pi /4} {\ln 2\,\,dx - I}\\ \\\,\, &\;\;= \frac{\pi }{4}\ln 2 - I\\\\\Rightarrow \quad 2I &\;\;= \frac{\pi }{4}\ln 2\\\\\Rightarrow \quad\;\;I &\;\;= \frac{\pi }{8}\ln 2\end{align}\]

Again, we see that using property - 9 saved us from the tedious task of first evaluating the anti-derivative.

Example –15

Evaluate \(\begin{align}I = \int\limits_0^{\pi /2} {\frac{{x\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx}\end{align}\)

Solution : We first use property -9 to simplify this integral; use the substitution \(\begin{align} x \to \frac{\pi }{2} - x\end{align}\) in the function to be integrated:

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

Adding the original and the modified expressions of I, we obtain :

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2I = \frac{\pi }{2}\int\limits_0^{\pi /2} {\frac{{\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}} \,\,dx\\&\;\Rightarrow \qquad\;\;\;\;\; I = \frac{\pi }{4}\int\limits_0^{\pi /2} {\frac{{\tan x{{\sec }^2}x}}{{1 + {{\tan }^4}x}}} \,\,dx\end{align}\]

Now we use the substitution \({\tan ^2}x = y\)

\[ \Rightarrow 2{\tan}x\;{\sec ^2}\;x\;\,\,dx = dy\]

                                                                        and when \(x = 0 \quad\Rightarrow \quad y = 0\)

\[\begin{align}&x = \frac{\pi }{2} \quad\Rightarrow\quad y = \infty\end{align} \]

 I can now be written in terms of y as :

\[\begin{align}&I = \frac{\pi }{8}\int\limits_0^\infty {\frac{{dy}}{{1 + {y^2}}}} \\&\left. {\,\,\,\, = \frac{\pi }{8}\left( {{{\tan }^{ - 1}}y} \right)} \right|_0^\infty \\\,\,\, &\;\;= \frac{{{\pi ^2}}}{{16}}\end{align}\]