A) \[\frac{1}{10}\left( \frac{\pi }{4}-{{\tan }^{-1}}\left( \frac{1}{9\sqrt{3}} \right) \right)\]
B) \[\frac{1}{20}{{\tan }^{-1}}\left( \frac{1}{9\sqrt{3}} \right)\]
C) \[\frac{1}{5}\left( \frac{\pi }{4}-{{\tan }^{-1}}\left( \frac{1}{3\sqrt{3}} \right) \right)\]
D) \[\frac{\pi }{40}\]
Correct Answer: A
Solution :
Let \[I=\int\limits_{\pi /6}^{\pi /4}{\frac{dx}{\sin 2x(ta{{n}^{5}}x+co{{t}^{5}}x)}}\] \[=\int\limits_{\pi /6}^{\pi /4}{\frac{{{\tan }^{5}}x}{2\sin x\cos x(ta{{n}^{10}}x+1)}}dx\] \[=\frac{1}{2}\int\limits_{\pi /6}^{\pi /4}{\frac{{{\tan }^{4}}x{{\sec }^{2}}x\,dx}{(1+{{\tan }^{10}}x)}}dx\] \[=\frac{1}{10}\int\limits_{{{(1/\sqrt{3})}^{5}}}^{1}{\frac{dt}{1+{{t}^{2}}}}\] [Putting \[{{\tan }^{5}}x=t\]] \[=\frac{1}{10}{{[ta{{n}^{-1}}t]}^{1}}_{{{(1/\sqrt{3})}^{5}}}\] \[=\frac{1}{10}\left[ \frac{\pi }{4}-{{\tan }^{-1}}\frac{1}{9\sqrt{3}} \right]\]
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