What is the value of the definite integral \[\int\limits_{ - a}^a {\sin xf\left( {\cos x} \right)dx} \]?
A. \[2\int\limits_0^a {\sin xf\left( {\cos x} \right)dx} \]
B. 0
C. 1
D. None of these

Hint: Here, a definite integral is given. First, check whether the function present in the given integral is odd or an even function. If the function is odd, then apply the property of the definite integral for the odd function. If the function is even, then apply the property of the definite integral for the even function and solve the integral to get the required answer.



Formula Used:\[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function. Means, \[f\left( { - x} \right) = - f\left( x \right)\]
\[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function. Means, \[f\left( { - x} \right) = f\left( x \right)\]



Complete step by step solution:The given definite integral is \[\int\limits_{ - a}^a {\sin xf\left( {\cos x} \right)dx} \].

Let consider,
\[g\left( x \right) = \sin xf\left( {\cos x} \right)\]
Now let’s calculate the value of \[g\left( { - x} \right)\].
\[g\left( { - x} \right) = \sin \left( { - x} \right)f\left( {\cos \left( { - x} \right)} \right)\]
Apply the trigonometric properties \[\sin \left( { - \theta } \right) = - \sin \theta \] and \[\cos \left( { - \theta } \right) = \cos \theta \].
\[ \Rightarrow g\left( { - x} \right) = - \sin f\left( {\cos x} \right)\]
\[ \Rightarrow g\left( { - x} \right) = - g\left( x \right)\]
Therefore, \[g\left( x \right) = \sin xf\left( {\cos x} \right)\] is an odd function.
Now apply the property of the definite integral \[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function.
We get,
\[\int\limits_{ - a}^a {\sin xf\left( {\cos x} \right)dx} = 0\]



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 whether the given trigonometric function is odd or even.