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Question: Integrate: \(\int_{-\pi /2}^{\pi /2}{{{e}^{{{\sin }^{-2}}x}}.{{\sin }^{2n+1}}x\ dx}=\)...

Integrate: ∫−π/2π/2esin⁡−2x.sin⁡2n+1x dx=\int_{-\pi /2}^{\pi /2}{{{e}^{{{\sin }^{-2}}x}}.{{\sin }^{2n+1}}x\ dx}=

Explanation

Solution

  1. If f(x)f(x) is an even function, then ∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^{a}{f(x)\ dx}=2\int_{0}^{a}{f(x)\ dx} .
    If f(x)f(x) is an odd function, then ∫−aaf(x) dx=0\int_{-a}^{a}{f(x)\ dx=0} .
  2. The expression sin⁡−2x{{\sin }^{-2}}x is equal to (sin⁡−1x)2{{\left( {{\sin }^{-1}}x \right)}^{2}} and sin⁡2n+1x{{\sin }^{2n+1}}x is equal to (sin⁡x)2n+1{{\left( \sin x \right)}^{2n+1}} .
  3. Use the property that sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta and sin⁡−1(−x)=−sin⁡−1x{{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x , to check if the expression to be integrated is an even expression or odd.

Complete step by step solution:
Let f(x)=esin⁡−2x.sin⁡2n+1xf(x)={{e}^{{{\sin }^{-2}}x}}.{{\sin }^{2n+1}}x . It can also be written as f(x)=e(sin⁡−1x)2.(sin⁡x)2n+1f(x)={{e}^{{{\left( {{\sin }^{-1}}x \right)}^{2}}}}.{{\left( \sin x \right)}^{2n+1}} for a better apprehension.
Let us find the expression for f(−x)f(-x) to compare and check whether it is an even function or odd.
f(−x)=e[sin⁡−1(−x)]2.[sin⁡(−x)]2n+1f(-x)={{e}^{{{\left[ {{\sin }^{-1}}(-x) \right]}^{2}}}}.{{\left[ \sin (-x) \right]}^{2n+1}}
Using the fact that sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta and sin⁡−1(−x)=−sin⁡−1x{{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x , we get:
⇒ f(−x)=e[−sin⁡−1x]2.[−sin⁡x]2n+1f(-x)={{e}^{{{\left[ -{{\sin }^{-1}}x \right]}^{2}}}}.{{\left[ -\sin x \right]}^{2n+1}}
⇒ f(−x)=e(−1)2(sin⁡−1x)2.(−1)2n+1(sin⁡x)2n+1f(-x)={{e}^{{{\left( -1 \right)}^{2}}{{\left( {{\sin }^{-1}}x \right)}^{2}}}}.{{\left( -1 \right)}^{2n+1}}{{\left( \sin x \right)}^{2n+1}}
Since 2n+12n+1 is odd for any value of n, we get:
⇒ f(−x)=e(sin⁡−1x)2.(−1)(sin⁡x)2n+1f(-x)={{e}^{{{\left( {{\sin }^{-1}}x \right)}^{2}}}}.\left( -1 \right){{\left( \sin x \right)}^{2n+1}}
⇒ f(−x)=−f(x)f(-x)=-f(x)
This means that f(x)f(x) is an odd function and therefore ∫−aaf(x) dx\int\limits_{-a}^{a}{f(x)\ dx} must be 0, for any real number a.

∴ ∫−π/2π/2esin⁡−2x.sin⁡2n+1x dx=0\int_{-\pi /2}^{\pi /2}{{{e}^{{{\sin }^{-2}}x}}.{{\sin }^{2n+1}}x\ dx}=0 is the answer.

Note:

  1. A function f(x) is:
    Even, if f(−x)=f(x)f(-x)=f(x) .
    Odd, if f(−x)=−f(x)f(-x)=-f(x) .
    None, in other cases.
  • An even/odd function is different from an even/odd number.
  1. (−x)even number=x{{(-x)}^{\text{even number}}}=x and (−x)odd number=−x{{(-x)}^{\text{odd number}}}=-x .
  2. Definite Integral: If ∫f(x)dx=g(x)+C\int{f\left( x \right)dx}=g\left( x \right)+C , then ∫abf(x)dx=[g(x)]ab=g(b)−g(a)\int\limits_{a}^{b}{f(x)dx}=[g(x)]_{a}^{b}=g(b)-g(a) .