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The integral\(\;\mathop \smallint \nolimits_{\pi /6}^{\pi /3} {\rm{se}}{{\rm{c}}^{2/3}}x{\rm{cose}}{{\rm{c}}^{4/3}}x\;dx{\rm{\;is}}\)equal to:1. 35⁄6 - 32⁄32. 34⁄3 - 31⁄33. 37⁄6 - 35⁄64. 35⁄3 - 31⁄3 |
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Answer» Correct Answer - Option 3 : 37⁄6 - 35⁄6 Let, \(I = \mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\rm{se}}{{\rm{c}}^{\frac{2}{3}}}x\cdot{\rm{cose}}{{\rm{c}}^{\frac{4}{3}}}\;xdx\) \( = \mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{{1.dx}}{{{\rm{co}}{{\rm{s}}^{\frac{2}{3}}}x\cdot{\rm{si}}{{\rm{n}}^{\frac{4}{3}}}x}}\) \(\left[\because {\frac{1}{{cos\theta }} = sec\;\theta \;and\;\frac{1}{{sin\theta }} = cosec\theta } \right]\) \( = \mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{{1dx}}{{{\rm{co}}{{\rm{s}}^2}x\cdot{\rm{ta}}{{\rm{n}}^{\frac{4}{3}}}x}} = \mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{{{\rm{se}}{{\rm{c}}^2}xdx}}{{{\rm{ta}}{{\rm{n}}^{\frac{4}{3}}}x}}\) Let tanx = u \(I = \mathop \smallint \nolimits_{\frac{1}{{\sqrt 3 }}}^{\sqrt 3 } {u^{ - \frac{4}{3}}}\;du\) \( = \frac{{\left[ {{u^{ - \frac{1}{3}}}} \right]_{\frac{1}{{\sqrt 3 }}}^{\sqrt 3 }}}{{ - \frac{1}{3}}} = \frac{{3\left[ {{u^{ - \frac{1}{3}}}} \right]_{\frac{1}{{\sqrt 3 }}}^{\sqrt 3 }}}{{ - 1}}\) \( = - 3\left[ {{3^{ - \;\frac{1}{6}}} - \frac{1}{{{3^{\frac{{ - 1}}{6}}}}}} \right] = - 3\left( {{3^{\frac{{ - 1}}{6}}} - {3^{\frac{1}{6}}}} \right) = 3\left( {{3^{\frac{1}{6}}} - {3^{\frac{{ - 1}}{6}}}} \right) = \left( {{3^{\frac{7}{6}}} - {3^{\frac{5}{6}}}} \right)\) |
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