Mathematics

Integrate $$\int _{ -\pi /2 }^{ \pi /2 }{ { e }^{ \sin ^{ -2 }{ x }  } } .\sin ^{ 2n+1 }{ x } dx=$$


SOLUTION
Let  $$I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{e^{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}}}{{\sin }^{2n + 1}}xdx} $$
  since  $$f\left( x \right) = {e^{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}}}{\sin ^{2n + 1}}xdx$$ is an odd function 
So, $$I = 0$$
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Subjective Medium Published on 17th 09, 2020
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