What is the value of the integral \[\int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}\left( {2n + 1} \right)xdx} \], for any integer \[n\]?
A. \[ - 1\]
B. 0
C. 1
D. \[\pi \]

Hint: Here, a definite integral is given. First, consider the term present in the given integral as \[f\left( x \right) = {e^{{{\sin }^2}x}}{\cos ^3}\left( {2n + 1} \right)x\]. Then calculate the value of \[f\left( {\pi - x} \right)\] by using the trigonometric properties. After that, substitute the values in the integration rule \[\int\limits_0^{2a} {f\left( x \right) dx} = \int\limits_0^a {f\left( x \right) dx} + \int\limits_0^a {f\left( {2a - x} \right) dx} \] and solve it to get the required answer.

Formula Used: \[{\sin ^n}\left( {\pi - \theta } \right) = \sin \theta \]
\[{\cos ^n}\left( {\pi - \theta } \right) = - {\cos ^n}\theta \], if \[n\] is odd number.
Integration Rule: \[\int\limits_0^{2a} {f\left( x \right) dx} = \int\limits_0^a {f\left( x \right) dx} + \int\limits_0^a {f\left( {2a - x} \right) dx} \]

Complete step by step solution: The given definite integral is \[\int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}\left( {2n + 1} \right)xdx} \].

Let consider,
\[f\left( x \right) = {e^{{{\sin }^2}x}}{\cos ^3}\left( {2n + 1} \right)x\]
Now let’s calculate the value of \[f\left( {\pi - x} \right)\].
\[f\left( {\pi - x} \right) = {e^{{{\sin }^2}\left( {\pi - x} \right)}}{\cos ^3}\left( {2n + 1} \right)\left( {\pi - x} \right)\]
\[ \Rightarrow f\left( {\pi - x} \right) = {e^{{{\sin }^2}\left( {\pi - x} \right)}}{\cos ^3}\left[ {\left( {2n + 1} \right)\pi - \left( {2n + 1} \right)x} \right]\]
Apply the trigonometric properties \[{\sin ^n}\left( {\pi - \theta } \right) = \sin \theta \] and \[{\cos ^n}\left( {\pi - \theta } \right) = - {\cos ^n}\theta \], if \[n\] is odd number.
\[ \Rightarrow f\left( {\pi - x} \right) = {e^{{{\sin }^2}x}}\left[ { - {{\cos }^3}\left( {2n + 1} \right)x} \right]\]
\[ \Rightarrow f\left( {\pi - x} \right) = - {e^{{{\sin }^2}x}}{\cos ^3}\left( {2n + 1} \right)x\]
\[ \Rightarrow f\left( {\pi - x} \right) = - f\left( x \right)\]

Now apply the rule of the definite integral \[\int\limits_0^{2a} {f\left( x \right) dx} = \int\limits_0^a {f\left( x \right) dx} + \int\limits_0^a {f\left( {2a - x} \right) dx} \]
We get,
\[\int\limits_0^\pi {f\left( x \right)dx} = \int\limits_0^{\dfrac{\pi }{2}} {f\left( x \right)dx} + \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\pi - x} \right)dx} \]
Substitute the values in the above integral equation.
\[\int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}\left( {2n + 1} \right)xdx} = \int\limits_0^{\dfrac{\pi }{2}} {f\left( x \right)dx} + \int\limits_0^{\dfrac{\pi }{2}} { - f\left( x \right)dx} \]
\[ \Rightarrow \int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}\left( {2n + 1} \right)xdx} = \int\limits_0^{\dfrac{\pi }{2}} {f\left( x \right)dx} - \int\limits_0^{\dfrac{\pi }{2}} {f\left( x \right)dx} \]
\[ \Rightarrow \int\limits_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}\left( {2n + 1} \right)xdx} = 0\]

Option ‘B’ is correct

Note: Sometimes students get confused and try to solve the integral by using the trigonometric identities. We can solve this integral by using the methods of indefinite integral. But the answer will be wrong.
So, to calculate the correct answer in the definite integral, first check the behaviour of the trigonometric functions for the different exponents and intervals.