Question

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

Answer 1

Hint: Put the value of \[x\] as \[ - x\] in the given integral to check whether given function odd function or even function. Then solve the integration.

Formula used:
\[\int_{ - a}^a {f\left( x \right)dx} = 0\] if \[f\left( x \right)\] is an odd function, [ i.e \[f\left( -x \right)=-f\left( x \right)\] ]

Complete step by step solution:
Given, the integral is:
\[I{\rm{ }} = \int\limits_{ - \pi }^\pi {{{\sin }^3}x{{\cos }^2}xdx} \]
 Assume that, \[ f \left(x\right) ={{\sin }^3}\left( { x} \right){{\cos }^2}\left( { x} \right) \]
Substitute \[x=-x\] in \[ f\left(x\right)\]
\[ f\left(-x\right) ={{\sin }^3}\left( { -x} \right){{\cos }^2}\left( { - x} \right) \]
\[ \Rightarrow f\left(-x\right) =-\sin^3\left( { x} \right)\cos ^2\left( { x} \right) \]
\[ \Rightarrow f\left(-x\right) =- f\left(x\right) \]
So that the function is an odd function.
Now apply the definite integral property: \[\int_{ - a}^a {f\left( x \right)dx} = 0\]
\[\int\limits_{ - \pi }^\pi {\sin }^3x{{\cos }^2}xdx =0\]

The correct answer is option D.

Additional information:
If an integration has upper limit and lower limit then the integration is known as definite integral. A definite integral is written as \[\int\limits_{a }^b f\left(x\right) dx\], where b is upper limit and a is lower limt of the integration.
Some properties of definite integral:
1. \[\int\limits_{a }^b f\left(x\right) dx = -\int\limits_{b }^a f\left(x\right) dx\]
2. \[\int\limits_{a }^a f\left(x\right) dx =0\]
Note: Students often do a common mistake to apply the formula of definite integrals. If a function is even function then the formula should be \[\int\limits_{-a }^a f\left(x\right) dx =2 \int\limits_{0 }^a f\left(x\right) dx\] , where \[f\left(-x\right) =f\left(x\right)\]. If a function is odd function then the formula should be \[\int\limits_{-a }^a f\left(x\right) dx =0\], where \[f\left(-x\right) = - f\left(x\right)\].