Find the value of \(\rm \int_{-\pi/2}^{\pi/2} sin^9x\ dx\)

Question

Question

Answer 1

Concept:

A function, f(x) is odd if f(-x) = - f(x) and the function is even if f(-x) = f(x)

If the function is even or odd and the interval is [-a, a], we can apply these rules:

When f(x) is even:

\(\rm \int_{-a}^{a} f(x)\ dx = 2\int_{0}^{a} f(x)\ dx\)

When f(x) is odd:

\(\rm \int_{-a}^{a} f(x)\ dx =0\)

Calculation:

The given function is, f(x) = sin⁹x

f(-x) = sin⁹(-x)

= - sin⁹x

= - f(x)

Since the function is odd the value of \(\rm \int_{-\pi/2}^{\pi/2} sin^9x\ dx\) = 0