Mathematics

$$\displaystyle\int^{\pi}_0\sqrt{|\cos x|-|\cos^3x|}dx$$ is?


ANSWER

$$\dfrac{4}{3}$$


SOLUTION
Given : $$\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right| -\left| \cos ^{ 3 }{ x }  \right| dx }  } $$
$$I=\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right| -\left| \cos ^{ 3 }{ x }  \right| dx }  } =\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right| \left| 1-\cos ^{ 2 }{ x }  \right| dx }  } \\ I=\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right|  } \left| \sin { x }  \right| dx } \\ I=\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right|  } \left( \sin { x }  \right) dx } \\ \sin { x } >0\quad (0<x<\pi )\\ \cos { x } =t\quad x\rightarrow 0\quad t\rightarrow 1\\ dt=-\sin { x } dx\quad x\rightarrow \pi \quad t\rightarrow -1\\ I=-\int _{ 1 }^{ -1 }{ \sqrt { \left| t \right|  } dt } \\ =\int _{ -1 }^{ 1 }{ \sqrt { \left| t \right|  } dt } \\ =\int _{ -1 }^{ 0 }{ \sqrt { -t } dt } +\int _{ 0 }^{ 1 }{ \sqrt { t } dt } \\ I=-\cfrac { 2 }{ 3 } \left| \left( -t \right) ^{ \cfrac { 3 }{ 2 }  } \right| _{ -1 }^{ 0 }+\cfrac { 2 }{ 3 } \left| t^{ \cfrac { 3 }{ 2 }  } \right| _{ 0 }^{ 1 }\\ I=-\cfrac { 2 }{ 3 } \left[ 0-1 \right] +\cfrac { 2 }{ 3 } (1-0)\\ =\cfrac { 4 }{ 3 } $$
Hence the correct answer is $$\cfrac { 4 }{ 3 } $$
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Single Correct Hard Published on 17th 09, 2020
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