# Definite Integration Set 3

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Example -5

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

Solution: By property - 9, we use the substitution $$x \to \pi - x$$ in the function to be integrated:

\begin{align}&I = \int\limits_0^\pi {\frac{{(\pi - x)\sin (2\pi - 2x)\sin \left( {\frac{\pi }{2}\cos \left( {\pi - x} \right)} \right)}}{{2(\pi - x) - \pi }}} \,\,dx\\\,\,\,\, &\;= \int\limits_0^\pi {\frac{{(\pi - x) \cdot ( - \sin 2x) \cdot \left( { - \sin \left( {\frac{\pi }{2}\cos x} \right)} \right)}}{{\pi - 2x}}} \,\,dx\\\,\,\,\, &\;= \int\limits_0^\pi {\frac{{(x - \pi )\sin 2x\sin \left( {\frac{\pi }{2}\cos x} \right)}}{{2x - \pi }}} \,\,dx\end{align}

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

\begin{align}&2I = \int\limits_0^\pi {\frac{{(2x - \pi )\sin 2x\sin \left( {\frac{\pi }{2}\cos x} \right)}}{{2x - \pi }}\,\,dx} \\\,\,\,\,\,\, &\;\;\;= \int\limits_0^\pi {\sin 2x\sin \left( {\frac{\pi }{2}\cos x} \right)\,\,dx} \end{align}

We use the substitution \begin{align}\frac{\pi }{2}\cos x = t\end{align}  in this simplified integral. The limits change from \begin{align}(0\,\,{\rm{to }}\;\pi )\end{align} to \begin{align}\left({\frac{\pi }{2}\,\,{\rm{to}}\frac{{ - \pi }}{2}} \right)\end{align} :

\begin{align}&2I = 2\int\limits_0^\pi {\sin x\cos x\sin \left( {\frac{\pi }{2}\cos x} \right)dx} \\\,\,\,\,\,\,\, &\;\;\;\;= \frac{{ - 8}}{{{\pi ^2}}}\int\limits_{\pi {\rm{/2}}}^{{\rm{ - }}\pi {\rm{/2}}} {t\sin t\,\,dt} \\\,\,\,\,\,\,\, &\;\;\;\;= \frac{8}{{{\pi ^2}}}\int\limits_{ - \pi {\rm{/2}}}^{\pi {\rm{/2}}} {t\sin t\,dt} \\\,\,\,\,\,\,\, &\;\;\;\;= \frac{8}{{{\pi ^2}}}\left\{ {\left. { - t\cos t} \right|_{ - \pi {\rm{2}}}^{\pi {\rm{/2}}} + \int\limits_{ - \pi {\rm{/2}}}^{\pi {\rm{/2}}} {\cos t\,dt} } \right\}\\\,\,\,\,\,\,\, &\;\;\;\;= \frac{8}{{{\pi ^2}}}\left\{ {0 + 2} \right\}\\\,\,\,\,\,\,\, &\;\;\;\;= \frac{{16}}{{{\pi ^2}}}\\ \Rightarrow\quad I &\;\;\;\;= \frac{8}{{{\pi ^2}}}\end{align}

Example -6

Evaluate \begin{align}I = \int\limits_{ - \pi }^\pi {\frac{{2x(1 + \sin x)}}{{1 + {{\cos }^2}x}}dx}\end{align}

Solution: The limits of integration for I are symmetric about x = 0; this suggests that we should look for even/odd portions present, if any, in the function to be integrated:

\begin{align}&I = 2\int\limits_\pi ^\pi {\frac{{\mathop {x\sin x}\limits^{\scriptstyle\,\,\,\,\,{\rm{even}}\atop\scriptstyle} }}{{1 + {{\cos }^2}x}}dx + 2\int\limits_{ - \pi }^\pi {\frac{{\mathop x\limits^{\scriptstyle{\rm{odd}}\atop\scriptstyle} }}{{1 + {{\cos }^2}x}}dx} } \\\,\,\,\, &\;= 4\int\limits_0^\pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}\,dx + 0} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\\\,\,\, &\;= 4\int\limits_0^\pi {\frac{{(\pi - x)\sin (\pi - x)}}{{1 + {{\cos }^2}(\pi - x)}}dx} \\\,\,\,&\; = 4\int\limits_0^\pi {\frac{{(\pi - x)\sin x}}{{1 + {{\cos }^2}x}}dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\end{align}

Adding the expressions for I in (1)and (2), we obtain:

$2I = 4\pi \int\limits_0^\pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}dx}$

Now use the substitution cos x = t; $$\sin x\,\,dx$$ becomes –dt and the limits of integration change from (0 to $$\pi$$ ) to (1 to –1):

\begin{align}&I = 2\pi \int\limits_1^{ - 1} {\frac{{ - dt}}{{1 + {t^2}}}} \,\,\,\,\\\,\,\, &\;= \left. {2\pi ({{\tan }^{ - 1}}t)} \right|_{ - 1}^1\\\,\,\,\, &\;= {\pi ^2}\end{align}

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