Splitting Integrand Into Two Functions

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(11) \(\int\limits_0^{2a} {f(x)dx = } \int\limits_0^a {\left\{ {f(x) + f(2a - x)} \right\}dx} \)

The justification for this property is described below:

\[\int\limits_0^{2a} {f(x)dx} = \int\limits_0^a {f(x)dx + } \int\limits_a^{2a} {f(x)dx} \]

To evaluate , \(\int\limits_a^{2a} {f(x)dx} \) we can equivalently use the variable \((2a - x)\) instead of x, but the limits of integration will change from (a to 2a) to (0 to a). This is because as x varies from 0 to a, 2ax will vary from (2a to a) covering the same interval [a, 2a]. Thus,

\[\int\limits_a^{2a} {f(x)dx = } \int\limits_0^a {f(2a - x)\,dx} \]

Hence, the stated assertion is valid

Example –17

If f is an even function, then prove that \(\int\limits_0^{\pi /2} {f(\cos 2x)\cos x\,dx = \sqrt 2 \,\int\limits_0^{\pi /4} {f(\sin 2x)\cos x\,dx} } \)

Solution: On the left side, the integration limits are \(\left( {0\,\,{\rm{to }}\frac{\pi }{2}} \right)\) while on the right side, they are \(\left( {0\,\,{\rm{to }}\frac{\pi }{4}} \right)\) .

Thus, it would be appropriate to use Property -11

\[\begin{align} \int\limits_{0}^{\pi /2}{f(\cos 2x)\cos x\,dx}&=\int\limits_{0}^{\pi /4}{\left\{ f\left( \cos 2x \right)\cos x+f\left( \cos 2\left( \frac{\pi }{2}-x \right) \right)\cos \left( \frac{\pi }{2}-x \right) \right\}\,dx} \\ & =\int\limits_{0}^{\pi /4}{\left\{ f(\cos 2x)\cos x+f(-\cos 2x)\sin x \right\}\,\,dx} \\ & ~~~[sincefis\text{ }an\text{ }even\text{ }function,\text{ }f(-\cos 2x)=f(\cos 2x)] \\ & =\int\limits_{0}^{\pi /4}{f(\cos 2x)\{\cos x+\sin x\}\,dx} \\ &=\sqrt{2}\int\limits_{0}^{\pi /4}{f(\cos 2x)\sin \left( x+\frac{\pi }{4} \right)\,\,dx}x \qquad \qquad \qquad \dots (1) \\&\left[ \begin{array}{l}{\rm{Now\;we\;use\;property\; - \;9\; to\;obtain\;the\;final\;form\;that\;we\;require;\;}}\\{\rm{use\;the\;substitution\;}}x \to \frac{\pi }{4}\, - x{\rm{\;in\;the\;function\;to\;be\;integrated\;in (1)}}\end{array} \right]\\ &= \sqrt 2 \int\limits_0^{\pi /4} {f\left( {\cos 2\left( {\frac{\pi }{4} - x} \right)} \right)\sin \left( {\frac{\pi }{4} - x + \frac{\pi }{4}} \right)\,dx} \\& = \sqrt 2 \int\limits_0^{\pi /4} {f(\sin 2x)\cos x\,\,dx} \end{align} \]

We could also have started with property -9 directly:

\[\begin{align}&I = \int\limits_0^{\pi /2} {f(\cos 2x)\cos x\,\,dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\\\,\,\,\, &= \int\limits_0^{\pi /2} {f\left( {\cos 2\left( {\frac{\pi }{2} - x} \right)} \right)\cos \left( {\frac{\pi }{2} - x} \right)\,dx} \\\,\,\,\, &= \int\limits_0^{\pi /2} {f\left( { - \cos 2x} \right)\sin x\,dx} \\\,\,\,\, &= \int\limits_0^{\pi /2} {f\left( {\cos 2x} \right)\sin x\,dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 3 \right)\end{align}\]

Adding (2) and (3) we obtain

\[2I = \int\limits_0^{\pi /2} {f(\cos 2x)(\sin x + \cos x)\,dx} \]

Notice now that the function being integrated on the right side above is symmetric about\(\begin{align}\frac{\pi }{4}\end{align};\,\) i.e., if we substitute \(\begin{align}\frac{\pi }{2} - x\end{align}\) for x, we will obtain the same function again. Thus, (property 7):

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

This is the same expression that we had obtained in (1). From here, we can proceed as described earlier.