JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer 90) \[\int_{\log \sqrt{\pi /2}}^{\log \sqrt{\pi }}{{{e}^{2x}}{{\sec }^{2}}\left( \frac{1}{3}{{e}^{2x}} \right)}dx\]is equal to:

    A) \[\sqrt{3}\]                                        

    B) \[\frac{1}{\sqrt{3}}\]

    C) \[\frac{3\sqrt{3}}{2}\]                                   

    D) \[\frac{1}{2\sqrt{3}}\]

    Correct Answer: A

    Solution :

    \[I=\int_{\log \sqrt{\pi /2}}^{\log \sqrt{\pi }}{{{e}^{2x}}\sec }\left( \frac{1}{3}{{e}^{2x}} \right)dx\] put\[{{e}^{2x}}=t\Rightarrow 2{{e}^{2x}}dx=dt\] \[x=\log \sqrt{\pi },\,\,t={{e}^{2\log \sqrt{\pi }}}=\pi \] When\[x=\log \sqrt{\pi /2},\,\,t={{e}^{2\log \sqrt{\pi /2}}}={{e}^{\log \pi /2}}=\frac{\pi }{2}\] When\[x=\log \sqrt{\pi },\,\,t={{e}^{2\log \sqrt{\pi }}}=\pi \] \[\therefore \]  \[I=\int_{\frac{\pi }{2}}^{\pi }{{{\sec }^{2}}\left( \frac{1}{3}t \right)dt}=\frac{1}{2}\cdot \frac{1}{\frac{1}{3}}\left[ \tan \frac{t}{3} \right]_{\pi /2}^{\pi }\] \[=\frac{3}{2}\left[ \tan \frac{\pi }{3}-\tan \frac{\pi }{6} \right]=\frac{3}{2}\left[ \sqrt{3}-\frac{1}{\sqrt{3}} \right]=\sqrt{3}\]

adversite



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