Mathematics

Solve:
$$\int\limits_0^{\pi /6} {\dfrac{{\cos 2x}}{{{{\left( {\cos x - \sin x} \right)}^2}}}dx} $$


ANSWER

$$ - \log \left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)$$


SOLUTION
$$\int _{ 0 }^{ \pi /6 }{ \dfrac { \cos  2x }{ \left( \cos  x-\sin  x \right) ^{ 2 } } dx } \\ \int _{ 0 }^{ \pi /6 }{ \dfrac { \cos ^{ 2 } x-\sin ^{ 2 } x }{ \left( \cos  x-\sin  x \right) ^{ 2 } } dx } \\ \int _{ 0 }^{ \pi /6 }{ \dfrac { (\cos  x-\sin  x)(\cos  x+\sin  x) }{ \left( \cos  x-\sin  x \right) ^{ 2 } } dx } \\ \int _{ 0 }^{ \pi/6 }{ \dfrac { (\cos  x+\sin  x) }{ (\cos  x-\sin  x) } dx } \\ put\quad t=\cos  x-\sin  x\\ Hence, dt=-(\cos  x+\sin  x)dx\\ -\displaystyle\int _{ 1 }^{ \frac { \sqrt { 3 } -1 }{ 2 }}{\dfrac {dt }{t}} \\ -\bigg[\log { t } \bigg]_{ 1 }^{ \dfrac { \sqrt { 3 } -1 }{ 2 }  }\\ -\left[ \log { \left( \dfrac { \sqrt { 3 } -1 }{ 2 }  \right) -\log { \left( 1 \right)  }  }  \right] \\ - \log { \left( \dfrac { \sqrt { 3 } -1 }{ 2 }  \right)  }   \quad \quad \quad \quad \because \log { \left( 1 \right) =0 } \\ \\ $$
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Single Correct Medium Published on 17th 09, 2020
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