Question

Evaluate the integral: \[ \int \sin^4 x \cos^4 x \, dx. \]

• A. \( \frac{1}{128} (-2 \sin^3 x \cos x - 3 \sin x \cos x + 3) + c \)
• B. \( \frac{1}{256} (-2 \sin^3 2x \cos 2x - 3 \sin 2x \cos 2x + 6x) + c \)
• C. \( \frac{1}{128} (2 \sin^3 x \cos x - 3 \sin x \cos x + 3x) + c \)
• D. \( \frac{1}{256} (3 \sin^3 x \cos x - 2 \sin x \cos x + 2) + c \)

Hint:

For integrals involving powers of \( \sin x \) and \( \cos x \), use double-angle and power-reduction identities to simplify the expression before integrating.

Answer 1

The Correct Option is B

Step 1: Expressing in Terms of Double Angles We use the double angle identity: \[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x. \] Rewriting the given integral: \[ I = \int \sin^4 x \cos^4 x \, dx = \int \left(\frac{1}{4} \sin^2 2x \right)^2 dx. \] Step 2: Substituting and Expanding Expanding: \[ I = \frac{1}{16} \int \sin^4 2x \, dx. \] Using the identity: \[ \sin^4 2x = \frac{3}{8} - \frac{1}{2} \cos 4x + \frac{1}{8} \cos 8x. \] Substituting: \[ I = \frac{1}{16} \int \left(\frac{3}{8} - \frac{1}{2} \cos 4x + \frac{1}{8} \cos 8x \right) dx. \] Step 3: Integrating Term by Term \[ I = \frac{1}{16} \left[ \frac{3}{8} x - \frac{1}{8} \sin 4x + \frac{1}{64} \sin 8x \right] + c. \] Simplifying: \[ I = \frac{1}{256} (-2 \sin^3 2x \cos 2x - 3 \sin 2x \cos 2x + 6x) + c. \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{\frac{1}{256} (-2 \sin^3 2x \cos 2x - 3 \sin 2x \cos 2x + 6x) + c}. \]