Mathematics

If $$x$$ satisfies the equation $$\displaystyle\left(\int_0^1{\frac{dt}{t^2+2t\cos{\alpha}+1}}\right)x^2-\left(\int_{-3}^3{\frac{t^2\sin{2t}}{t^2+1}dt}\right)x-2=0$$ 
for $$(0<\alpha<\pi)$$
then the value of $$x$$ is?


ANSWER

$$\displaystyle\pm2\sqrt{\frac{\sin{\alpha}}{\alpha}}$$


SOLUTION
$$\displaystyle \left( \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha  } +1 }  }  \right) x^{ 2 }-\left( \int _{ -3 }^{ 3 }{ \frac { t^{ 2 }\sin { 2t }  }{ t^{ 2 }+1 } dt }  \right) x-2=0$$

Here, $$\displaystyle \int _{ -3 }^{ 3 }{ \frac { t^{ 2 }\sin { 2t }  }{ t^{ 2 }+1 } dt } =0 \left(\because f(t)=\frac { t^{ 2 }\sin { 2t }  }{ t^{ 2 }+1 } \text{ is odd}\right)$$

So, the equation becomes
$$\displaystyle \left( \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha  } +1 }  }  \right) x^{ 2 }-2=0$$   ....(1)

Now,
$$\displaystyle  \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha  } +1 }  } =\int _{ 0 }^{ 1 }{ \frac { dt }{ (t+\cos { \alpha  } )^{ 2 }+\sin ^{ 2 }{ \alpha  }  }  } $$

$$=\displaystyle \frac { 1 }{ \sin { \alpha  }  } \left[{ \tan ^{ -1 }{ \left(\frac { t+\cos { \alpha} }{ \sin { \alpha  }  } \right) }  }\right]_{ 0 }^{ 1 }$$

$$\displaystyle =\frac { 1 }{ \sin { \alpha  }  }  \left[\tan ^{ -1 }{\cot {\dfrac{\alpha}2 } -\tan ^{ -1 }{ \cot { \alpha  } }  }\right]$$

$$\displaystyle =\frac { 1 }{ \sin { \alpha  }  } \left[\frac { \pi  }{ 2 }-\frac { \alpha  }{ 2 } -\frac { \pi  }{ 2 } +\alpha\right]$$

$$\displaystyle =\frac { \alpha  }{ 2\sin { \alpha  }  } $$

So, equation (1) becomes
$$\displaystyle\frac { \alpha  }{ 2\sin { \alpha  }  } x^{ 2 }-2=0$$

$$\Rightarrow \displaystyle x=\pm 2 \sqrt {\frac{\sin \alpha}{\alpha}}$$

Hence, option D.
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Single Correct Hard Published on 17th 09, 2020
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