MCQ
$\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\frac{{{x^2}}}{{1\, + \,\tan \,x\, + \,\sqrt {1 + {{\tan }^2}x} }}} \,dx$ મેળવો.
- A$\pi^3$
- B$\frac{{{\pi ^3}}}{{12}}$
- C$\frac{{{\pi ^3}}}{{24}}$
- D$\frac{{{\pi ^3}}}{{48}}$
$ = \int\limits_0^{\frac{\pi }{2}} {\frac{{{x^2}\left[ {2 + 2\sqrt {1 + {{\tan }^2}x} } \right]}}{{{{\left( {1 + \sqrt {1 + {{\tan }^2}x} } \right)}^2} - {{\tan }^2}x}}} dx$
$ = \int\limits_0^{\frac{\pi }{2}} {\frac{{2{x^2}\left( {1 + \sqrt {1 + {{\tan }^2}x} } \right)}}{{1 + 1 + {{\tan }^2}x + 2\sqrt {1 + {{\tan }^2}x} - {{\tan }^2}x}}} dx$
$ = \int\limits_0^{\frac{\pi }{2}} {\frac{{2{x^2}\left( {1 + \sqrt {1 + {{\tan }^2}x} } \right)}}{{2\left( {1 + \sqrt {1 + {{\tan }^2}x} } \right)}}} dx$
$=\left(\frac{x^{3}}{3}\right)_{0}^{\pi / 2} \Rightarrow \frac{\pi^{3}}{24}$
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વિધાન $-1$ : સમીકરણના ઉકેલની સંભાવના $1$ છે. .
વિધાન $-2$ : સમીકરણના અનન્ય ઉકેલની સંભાવના $\frac {3}{8}$ છે. .