Mathematics

Evaluate the following integral:
$$\int { \cfrac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } }  }  } dx\quad $$


SOLUTION
Consider $$I=\displaystyle\int { \dfrac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } }  }  } dx$$

$$=\displaystyle\int { \dfrac { 1 }{ \sqrt { -\left( { x }^{ 2 }-3x-8 \right)  }  }  } dx$$

$$=\displaystyle\int { \dfrac { 1 }{ \sqrt { -\left( { x }^{ 2 }-3x+\cfrac { 9 }{ 4 } -\dfrac { 9 }{ 4 } -8 \right)  }  }  } dx$$

$$\Rightarrow I=\displaystyle\int { \dfrac { 1 }{ \sqrt { -\left\{ { \left( x-\cfrac { 3 }{ 2 }  \right)  }^{ 2 }-{ \left( \dfrac { \sqrt { 41 } }{ 2 }  \right)  }^{ 2 } \right\}  }  }  } dx$$

$$=\displaystyle\int { \dfrac { 1 }{ \sqrt { { \left( \dfrac { \sqrt { 41 }  }{ 2 }  \right)  }^{ 2 }-{ \left( x-\cfrac { 3 }{ 2 }  \right)  }^{ 2 } }  }  } dx$$

We know that $$\displaystyle\int{\dfrac{dx}{\sqrt{{a}^{2}-{x}^{2}}}}=\dfrac{1}{a}{\sin}^{-1}{\dfrac{x}{a}}+c$$

Replace $$a\rightarrow\dfrac{\sqrt{41}}{2}$$ and $$x\rightarrow\,2x-3$$

$$=\sin ^{ -1 }{ \left( \dfrac { 2x-3 }{ \sqrt { 41 }  }  \right)  } +C$$

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Subjective Medium Published on 17th 09, 2020
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