The integral $\mathop \smallint \nolimits_{\pi /6}^{\pi /4} \frac{{dx}}{{{\rm{sin\;}}2x\left( {{\rm{ta}}{{\rm{n}}^5}x + {\rm{co}}{{\rm{t}}^5}x} \right)}}$ equals:

Question

The integral $\mathop \smallint \nolimits_{\pi /6}^{\pi /4} \frac{{dx}}{{{\rm{sin\;}}2x\left( {{\rm{ta}}{{\rm{n}}^5}x + {\rm{co}}{{\rm{t}}^5}x} \right)}}$ equals:

TuteeHUB · Accepted Answer

$\frac{\pi }{{40}}$

TuteeHUB · Answer

$\frac{1}{{20}}{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{1}{{9\sqrt 3 }}} \right)$

TuteeHUB · Answer

$\frac{1}{{10}}\left( {\frac{\pi }{4} - {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{1}{{9\sqrt 3 }}} \right)} \right)$

TuteeHUB · Answer

$\frac{1}{5}\left( {\frac{\pi }{4} - {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{1}{{3\sqrt 3 }}} \right)} \right)$

1.	The integral \(\mathop \smallint \nolimits_{\pi /6}^{\pi /4} \frac{{dx}}{{{\rm{sin\;}}2x\left( {{\rm{ta}}{{\rm{n}}^5}x + {\rm{co}}{{\rm{t}}^5}x} \right)}}\) equals:
A.	\(\frac{1}{{20}}{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{1}{{9\sqrt 3 }}} \right)\)
B.	\(\frac{1}{{10}}\left( {\frac{\pi }{4} - {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{1}{{9\sqrt 3 }}} \right)} \right)\)
C.	\(\frac{\pi }{{40}}\)
D.	\(\frac{1}{5}\left( {\frac{\pi }{4} - {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{1}{{3\sqrt 3 }}} \right)} \right)\)
Answer» C. \(\frac{\pi }{{40}}\)

The integral \(\mathop \smallint \nolimits_{\pi /6}^{\pi /4} \frac{{dx}}{{{\rm{sin\;}}2x\left( {{\rm{ta}}{{\rm{n}}^5}x + {\rm{co}}{{\rm{t}}^5}x} \right)}}\) equals:

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