ધારો કે $u = \frac{{{{\tan }^{ - 1}}x}}{{1 + {{\tan }^{ - 1}}x}}$ તથા $v = {\tan ^{ - 1}}x$
$\therefore u = \frac{v}{{1 + v}}$
હવે, $v$ ની સાપેક્ષ વિકલન કરતાં,
$\frac{{du}}{{dv}} = \frac{{\left( {1 + v} \right)\left( 1 \right) - v\left( 1 \right)}}{{{{\left( {1 + v} \right)}^2}}}$
$ = \frac{{1 + v - v}}{{{{\left( {1 + v} \right)}^2}}}$
$ = \frac{1}{{{{\left( {1 + v} \right)}^2}}} = \frac{1}{{{{\left( {1 + {{\tan }^{ - 1}}x} \right)}^2}}}$
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$f(x)=\max \{\sin t: 0 \leq t \leq x\}, \quad 0 \leq x \leq \pi$
$\quad \quad \quad \quad \quad \quad 2+\cos x,\quad \quad \quad \quad x>\pi$
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