Evaluating the Definite Integral \(\int_0^\pi {{\sin }^6}} x \cdot {\cos ^5}x dx\)

We are asked to find the value of the definite integral given by \(\int_0^\pi {{\sin }^6}} x \cdot {\cos ^5}x dx\). This involves integrating trigonometric functions over a specific interval.

Understanding the Integrand and Integration Limits

The integrand function is \(f(x) = {{\sin }^6}} x \cdot {\cos ^5}x\). The limits of integration are from \(0\) to \(\pi\).

To evaluate definite integrals with limits \(0\) to \(a\) or \(0\) to \(2a\), we can often use properties of definite integrals. A particularly useful property for the interval \(0\) to \(\pi\) (which can be seen as \(0\) to \(2 \times \pi/2\)) is related to the behavior of the function when \(x\) is replaced by \(\pi - x\).

Using Properties of Definite Integrals

Consider the property for definite integrals:

• If \(f(a - x) = f(x)\), then \(\int_0^a f(x) dx = 2\int_0^{a/2} f(x) dx\).
• If \(f(a - x) = -f(x)\), then \(\int_0^a f(x) dx = 0\).

In our case, the upper limit is \(\pi\), so we should check the property for \(a = \pi\).

Let's examine the integrand \(f(x) = {{\sin }^6}} x \cdot {\cos ^5}x\) at \(x = \pi - x\):

\(f(\pi - x) = {{\sin }^6}(\pi - x) \cdot {\cos ^5}(\pi - x)\)

We know the following trigonometric identities:

• \(\sin(\pi - x) = \sin x\)
• \(\cos(\pi - x) = -\cos x\)

Substituting these identities into the expression for \(f(\pi - x)\):

\(f(\pi - x) = (\sin x)^6 \cdot (-\cos x)^5\)

Since \((\sin x)^6 = {{\sin }^6}} x\) (an even power) and \((-\cos x)^5 = -{\cos ^5}x\) (an odd power preserves the negative sign), we get:

\(f(\pi - x) = {{\sin }^6}} x \cdot (-{\cos ^5}x)\)

\(f(\pi - x) = -{{\sin }^6}} x \cdot {\cos ^5}x\)

Comparing this with \(f(x)\), we see that \(f(\pi - x) = -f(x)\).

Applying the Zero Property for Definite Integrals

Since we found that \(f(\pi - x) = -f(x)\) for the definite integral \(\int_0^\pi f(x) dx\), we can apply the property that states if \(f(a-x) = -f(x)\) over the interval \([0, a]\), then \(\int_0^a f(x) dx = 0\).

Here, \(a = \pi\), and \(f(\pi - x) = -f(x)\). Therefore, the value of the definite integral is \(0\).

Thus, \(\int_0^\pi {{\sin }^6}} x \cdot {\cos ^5}x dx = 0\).

This property of definite integrals simplifies the calculation significantly, allowing us to determine the value without performing the explicit integration.