Question

Evaluate the integral using the properties of even and odd functions as an aid. $$ \int_{-\pi / 2}^{\pi / 2} \sin ^{2} x \cos x d x $$

Short answer

0

Step-by-step solution

Identify the type of the function

Check if the function \( \sin^2x \cos x \) is odd or even. An odd function is one that changes sign when its argument is changed to its negative. That is, if f(-x) = -f(x) for every x in the function's domain then the function is odd. \( \sin^2x \cos x \) is an odd function because \( \sin^2(-x) \cos(-x) = -\sin^2x \cos x \).

Apply the properties of odd functions to the integral

If a function is odd, the area under the function from -a to a is 0 because the area above the x-axis (-a to 0) is exactly balanced by the area below the x-axis (0 to a). The definite integral of an odd function over a symmetric interval is therefore always zero.

Solve the integral

Based on the property of odd functions, the integral \( \int_{-\pi / 2}^{\pi / 2} \sin ^{2} x \cos x d x \) equals to zero