Definite Integration Set 1

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Example - 1

Evaluate \(\begin{align}&\int\limits_0^{n/2} {\frac{{x\sin x\cos x}}{{{{({a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x)}^2}}}dx}\end{align}\) where \(\ a, b > 0\)

Solution: Consider only the function \(\begin{align}&\ f(x) = \frac{{\sin x\;\cos x}}{{{{({a^2}\;{{\cos }^2}x + {b^2}\;{{\sin }^2}x)}^2}}}\end{align}\)
Observe that \(f(x)\) can be easily integrated. Using the substitution \({\sin ^2}x = t,\) we get the integral as (verify):

\[\int {f(x)dx = \frac{{ - 1}}{{2({b^2} - {a^2})({a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x)}}} \]

Now since we know the integral of \(\ f (x)\), we use integration by parts on the original integral, which is nothing but:

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

Applying integration by parts, we obtain:

\[\begin{align}&I = \left. {\left( {x\int {f(x)dx} } \right)} \right|_0^{\pi {\rm{/}}2} - \int\limits_0^{\pi {\rm{/}}2} {\left( {\int {f(x)\,dx} } \right)} \,dx\\\,\, &\;= \left. {\frac{1}{{2({b^2} - {a^2})}}\left\{ {\frac{{ - x}}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}} \right.} \right|_0^{\frac{\pi }{2}}\left. { + \int\limits_0^{\pi {\rm{/}}2} {\frac{1}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}dx} } \right\}\\\,& = \frac{1}{{2({b^2} - {a^2})}}\left\{ {\frac{{ - \pi }}{{2{b^2}}} + \int\limits_0^{\pi {\rm{/}}2} {\frac{{{{\sec }^2}x}}{{{a^2} + {b^2}{{\tan }^2}x}}dx} } \right\}\\& \qquad \qquad \qquad \qquad\qquad\qquad\qquad\downarrow \\ & \qquad \qquad \qquad \left\{ \begin{array}{l}{\rm{To\;evaluate\;this\;integral,\;use\;the\;substitution }}\tan x = t{\rm{;}}\\{\rm{this\; integral\; reduces\; to }}\int\limits_0^\infty {\frac{{dt}}{{{a^2} + {b^2}{t^2}}}} \,\,\,{\rm{which}}\,\,{\rm{is\,}}\frac{\pi }{{2ab}}\end{array} \right\}\\& = \frac{1}{{2({b^2} - {a^2})}}\left\{ {\frac{\pi }{{2ab}} - \frac{\pi }{{2{b^2}}}} \right\}\\&= \frac{\pi }{{4a{b^2}(a + b)}}\end{align}\]

Example -2

Evaluate \(I = \int\limits_0^1 {\ln \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)dx} \)

Solution: The expression for I contains both \((1 - x)\) and \((1 + x);\) the substitution \(x = \cos 2\theta \) could do the job.

\[x = \cos 2\theta\\\Rightarrow\quad\ dx = - 2\sin 2\;\theta\; d\;\theta\\\Rightarrow\quad \ \text{when}\; x = 0, \theta = \pi /4\\ \;\;\;\;\;\;\;\text{when}\; x = 1, \theta = 0.\]

Thus, I gets modified to

\[\begin{align}&I = - 2\int\limits_{\pi {\rm{/}}4}^0 {\ln \left( {\sqrt {1 - \cos 2\theta } + \sqrt {1 + \cos 2\theta } } \right)\sin 2\theta d\theta } \\\,\,\,\, &\;= 2\int\limits_0^{\pi {\rm{/}}4} {\ln \left\{ {\sqrt 2 \left( {\sin \theta + \cos \theta } \right)} \right\}\sin 2\theta d\theta } \\\,\,\,\, &\;= 2\int\limits_0^{\pi {\rm{/}}4} {\left( {\ln \sqrt 2 } \right)\sin 2\theta d\theta + 2\int\limits_{\rm{0}}^{\pi {\rm{/4}}} {\ln (\mathop {\sin \theta }\limits_{\scriptstyle{\rm{First}}\atop\scriptstyle{\rm{Function}}} + \cos \theta )\mathop {\sin 2\theta }\limits_{\scriptstyle{\rm{Second}}\atop\scriptstyle{\rm{Function}}} d\theta } } \\\,\,\,\, &\;= \left. { - \ln \sqrt 2 (\cos 2\theta )} \right|_0^{\pi {\rm{/}}4} + 2\left\{ {\left. {\frac{{ - \ln (\sin \theta + \cos \theta )\cos 2\theta }}{2}} \right|_0^{\pi {\rm{/4}}} + \int\limits_0^{\pi /4} {\frac{{\cos \theta - \sin \theta }}{{\sin \theta + \cos \theta }} \cdot \frac{{\cos 2\theta }}{2}d\theta } } \right\}\\\,\,\,\, &\;= \ln \sqrt 2 + 2\left\{ {0 + \frac{1}{2}\int\limits_0^{\pi {\rm{/4}}} {{{(\cos \theta - \sin \theta )}^2}d\theta } } \right\}\\\,\,\,\, &\;= \ln \sqrt 2 + \int\limits_0^{\pi {\rm{/4}}} {{{(\cos \theta - \sin \theta )}^2}d\theta } \\\,\,\,\, &\;= \ln \sqrt 2 + \int\limits_0^{\pi {\rm{/4}}} {(1 - \sin 2\theta )d\theta } \\\,\,\,\, &\;= \ln \sqrt 2 \left. { + \frac{\pi }{4} + \frac{{\cos 2\theta }}{2}} \right|_0^{\pi {\rm{/4}}}\\\,\,\,\, &\;= \ln \sqrt 2 + \frac{\pi }{4} - \frac{1}{2}\end{align}\]

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