Question

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

Answer 1

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

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

∴ $\int_{-\pi /2}^{\pi /2}{{{e}^{{{\sin }^{-2}}x}}.{{\sin }^{2n+1}}x\ dx}=0$ is the answer.

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