Mathematics

Solve :$$\displaystyle \int_0^{\pi/2}\dfrac{dx}{1+\tan^3 x}$$ ?


ANSWER

$$\dfrac{\pi}{4}$$


SOLUTION
$$\quad \quad \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \dfrac { dx }{ 1+{ \tan  }^{ 3 }x }  } \\ =\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \dfrac { 1 }{ \left( 1+{ u }^{ 2 } \right) \left( 1+{ u }^{ 3 } \right)  } du } \quad \quad \left[ let,\quad u=\tan x \right] \\ =\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \dfrac { 1-2u }{ 3\left( { u }^{ 2 }-u+1 \right)  }  } +\dfrac { u+1 }{ 2\left( { u }^{ 2 }+1 \right)  } +\dfrac { 1 }{ 6\left( u+1 \right)  } du\\ =\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \dfrac { 1-2u }{ 3\left( { u }^{ 2 }-u+1 \right)  } du } +\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \dfrac { u+1 }{ 2\left( { u }^{ 2 }+1 \right)  } du } +\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \dfrac { 1 }{ 6\left( u+1 \right)  } du } \\ ={ \left[ -\dfrac { 1 }{ 3 } log\left[ { u }^{ 2 }-u+1 \right]  \right]  }_{ 0 }^{ \dfrac { \pi  }{ 2 }  }+{ \left[ \frac { 1 }{ 2 } \left( \dfrac { 1 }{ 2 } log\left( { u }^{ 2 }+1 \right) +\tan ^{ -1 }{ \left( u \right)  }  \right)  \right]  }_{ 0 }^{ \dfrac { \pi  }{ 2 }  }+{ \left[ \dfrac { 1 }{ 6 } log\left( u+1 \right)  \right]  }_{ \left( 0 \right)  }^{ \dfrac { \pi  }{ 2 }  }\\ =\dfrac { \pi  }{ 4 } -0\\ =\dfrac { \pi  }{ 4 } $$
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Single Correct Medium Published on 17th 09, 2020
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