Examples On Integration By Substitution Set-13

Example - 34

Evaluate \(\begin{align}\int {\frac{{\sqrt {\cos 2x} }}{{\sin x}}} \,\,dx\end{align}\)

Solution: \[\;\;\;\; I = \int {\frac{{\sqrt {{{\cos }^2}x - {{\sin }^2}x} }}{{\sin x}}} \,\,dx\\ = \int {\sqrt {{{\cot }^2}x - 1} } \,\,dx\]

The substitution \(\cot x = \sec \theta \) can help us simplify this integral:

\[\begin{align}&\qquad \cot x = \sec \theta \\ & \Rightarrow\quad - {\rm{cose}}{{\rm{c}}^2}xdx = \sec \theta \tan \theta d\theta \\ & \Rightarrow\quad I = \int {\sqrt {{{\sec }^2}\theta - 1} } \,\, \cdot \;\;\frac{{\sec \theta \tan \theta }}{{ - {\rm{cose}}{{\rm{c}}^2}x}}\,\,d\theta  \\ &\qquad\quad \begin{array}{l} = - \int {\frac{{\sec \theta {{\tan }^2}\theta }}{{1 + {{\sec }^2}\theta }}} \,\,d\theta \left\{ \begin{array}{l}{\rm{\;Now\;convert \;this\;expression\;to\;one\;}}\\{\rm{involving\;sin}}\theta \,{\rm{\;and\;cos}}\theta \,{\rm{\;terms}}\end{array} \right\}\\ = - \int {\frac{{{{\sin }^2}\theta }}{{\cos \theta + {{\cos }^3}\theta }}} \,\,d\theta \\ = - \int {\frac{{{{\sin }^2}\theta }}{{\cos \theta \left( {1 + {{\cos }^2}\theta } \right)}}} \,\,d\theta \left\{ \begin{array}{l}{\rm{Now}}\,{\rm{write si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}\\{\rm{as }}1 - {\cos ^2}\theta \end{array} \right\}\\ = \int {\frac{{{{\cos }^2}\theta - 1}}{{\cos \theta \left( {1 + {{\cos }^2}\theta } \right)}}} \,\,d\theta \\ = - \int {\frac{{\left( {{{\cos }^2}\theta + 1} \right) - 2{{\cos }^2}\theta }}{{\cos \theta \left( {1 + {{\cos }^2}\theta } \right)}}} \,\,d\theta \\ = - \int {\sec \theta d\theta + 2} \int {\frac{{\cos \theta }}{{1 + {{\cos }^2}\theta }}} \,\,d\theta \\ = - \ln \left| {\sec \theta + \tan \theta } \right| + 2\int {\frac{{\cos \theta }}{{2 - {{\sin }^2}\theta }}\,\,d\theta }\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \searrow \\ {\rm{Now\;use\;the\;substitution\;}}\sin \theta = t\\ \\ = - \ln \left| {\sec \theta + \tan \theta } \right| + 2\int {\frac{{dt}}{{{{\left( {\sqrt 2 } \right)}^2} - {t^2}}}} \\ = - \ln \left| {\sec \theta + \tan \theta } \right| + \frac{2}{{2\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 + t}}{{\sqrt 2 - t}}} \right| + C\\ = - \ln \left| {\cot x + \sqrt {{{\cot }^2}x - 1} } \right| + \frac{1}{{\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 + \sqrt {1 - {{\tan }^2}x} }}{{\sqrt 2 - \sqrt {1 - {{\tan }^2}x} }}} \right| + C\end{array}\ \end{align}\]
 

Example – 35

Evaluate \(\begin{align}\int {{{\sin }^{ - 1}}\left( {\frac{{2x + 2}}{{\sqrt {4{x^2} + 8x + 13} }}} \right)} \,\,dx\end{align}\)

Solution: We have to some how manipulate the expression inside the brackets so that it assumes the form \(\sin \theta .\) Why? Because that would lead to a cancellation of \('{\sin ^{ - 1}}'\,\,{\rm{with}}\,\,'\sin '\) and we’d be left with only \(\theta \) .

The denominator is

\[4{x^2} + 8x + 13\]

\[\begin{array}{l} = 4{x^2} + 8x + 4 + 9\\ = {\left( {2x + 2} \right)^2} + {3^2}\end{array}\]

We can use the substitution \(2x + 2 = 3\tan \theta \) to reduce the denominator to a (simpler) trignometric term:

\[\begin{align}&\qquad\;\; 2x + 2 = 3\tan \theta \\ &\Rightarrow\quad 2dx = 3{\sec ^2}\theta d\theta \\ &\Rightarrow\quad I= \int {{{\sin }^{ - 1}}\left( {\frac{{3\tan \theta }}{{\sqrt {{3^2}{{\tan }^2}\theta } + {3^2}}}} \right)} \cdot \left( {\frac{3}{2}{{\sec }^2}\theta d\theta } \right) \\ &\quad\qquad\;= \frac{3}{2}\int {{{\sin }^{ - 1}}\left( {\frac{{\tan \theta }}{{\sec \theta }}} \right)} {\sec ^2}\theta d\theta \\&\quad\qquad\; = \frac{3}{2}\int {{{\sin }^{ - 1}}\left( {\sin \theta } \right)} {\sec ^2}\theta d\theta \\&\quad\qquad\; = \frac{3}{2}\int \theta {\sec ^2}\theta d\theta \end{align}\]

We will continue the solution in example 42 when we’ve studied integration by parts. This example was included here to illustrate the substitution technique involved.