Question

Evaluate \( \int_{-\pi/2}^{\pi/2} \sin^9 x \cos^2 x \, dx \):

• A. \( \frac{2}{3} \)
• B. \( 1 \)
• C. \( \frac{1}{11} \)
• D. \( \frac{7\pi}{6} \)
• E. \( 0 \)

Hint:

When integrating odd functions over symmetric limits, the integral evaluates to zero. Recognizing symmetry can simplify the evaluation process.

Answer 1

Answer: (E)

We are tasked with evaluating the integral: \[ \int_{-\pi/2}^{\pi/2} \sin^9 x \cos^2 x \, dx \] First, observe that the integrand consists of powers of sine and cosine. 
We can use the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \) to simplify the integral: \[ \int_{-\pi/2}^{\pi/2} \sin^9 x \cos^2 x \, dx = \int_{-\pi/2}^{\pi/2} \sin^9 x \left( \frac{1 + \cos(2x)}{2} \right) \, dx \] 
Now, split the integral: \[ = \frac{1}{2} \int_{-\pi/2}^{\pi/2} \sin^9 x \, dx + \frac{1}{2} \int_{-\pi/2}^{\pi/2} \sin^9 x \cos(2x) \, dx \] The first integral is an odd function \( \sin^9 x \), and when integrated over symmetric limits from \( -\pi/2 \) to \( \pi/2 \), it evaluates to 0. 
The second integral involves \( \sin^9 x \cos(2x) \), which is also an odd function, so it too evaluates to 0. 
Therefore, the total integral evaluates to 0.