Question

Evaluate \[ \int \sin^3 x \cos^2 x \, dx = ? \]

• A. \(\frac{\sin^4 x}{5} - \frac{\sin^2 x \cos^2 x}{15} + 2 \cos x + c\)
• B. \(\frac{\sin^4 x}{5} - \frac{\sin^2 x \cos^2 x}{15} + 2 \cos x + c\)
• C. \(\frac{\sin^4 x}{5} - \frac{\sin^2 x \cos^2 x}{15} + 2 \cos x + c\)
• D. \(\frac{\sin^4 x}{5} - \frac{\sin^2 x \cos^2 x}{15} + 2 \cos x + c\)

Hint:

Use trigonometric identities and reduction formulas for integrals involving powers of sine and cosine.

Answer 1

The Correct Option is A

To solve this integral, we can use the reduction formula or simplify the powers of sine and cosine to reduce the integral into simpler parts. Here's how: 1. Express \(\sin^3 x \cos^2 x\) as \(\sin x (\sin^2 x \cos^2 x)\). 2. Use the identity \(\sin^2 x = 1 - \cos^2 x\) to further simplify the integral. 3. Split the integral into manageable parts and integrate term by term. The final solution gives us: \[ \frac{\sin^4 x}{5} - \frac{\sin^2 x \cos^2 x}{15} + 2 \cos x + c. \]