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What is the value of the integral \[\int\limits_{ - 1}^1 {{{\sin }^3}x{{\cos }^2}xdx} \]?
A. 0
B. 1
C. \[\dfrac{1}{2}\]
D. 2

Answer
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Hint: Here, a definite integral is given. First, check whether the function present in the given integral is an odd or an even function by calculating the value of \[f\left( { - x} \right)\]. 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 \[\int\limits_{ - 1}^1 {{{\sin }^3}x{{\cos }^2}xdx} \].

Let consider,
\[f\left( x \right) = {\sin ^3}x{\cos ^2}x\]
Now let’s calculate the value of \[f\left( { - x} \right)\].
\[f\left( { - x} \right) = {\sin ^3}\left( { - x} \right){\cos ^2}\left( { - x} \right)\]
\[ \Rightarrow f\left( { - x} \right) = - {\sin ^3}x{\cos ^2}x\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
Therefore, \[f\left( x \right)\] 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,
\[\int\limits_{ - 1}^1 {{{\sin }^3}x{{\cos }^2}xdx} = 0\]

Option ‘A’ 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.