Integral Sin Cos 9237F6

1. **State the problem:** We need to evaluate the integral $$\int 6 \sin 2x \cos x \, dx$$. 2. **Recall relevant formulas:** Use the product-to-sum identity for sine and cosine: $$\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$$ 3. **Apply the identity:** $$6 \sin 2x \cos x = 6 \times \frac{1}{2} [\sin(2x + x) + \sin(2x - x)] = 3 [\sin 3x + \sin x]$$ 4. **Rewrite the integral:** $$\int 6 \sin 2x \cos x \, dx = \int 3 [\sin 3x + \sin x] \, dx = 3 \int \sin 3x \, dx + 3 \int \sin x \, dx$$ 5. **Integrate each term:** - $$\int \sin 3x \, dx = -\frac{1}{3} \cos 3x + C$$ - $$\int \sin x \, dx = -\cos x + C$$ 6. **Combine results:** $$3 \times \left(-\frac{1}{3} \cos 3x\right) + 3 \times (-\cos x) + C = -\cos 3x - 3 \cos x + C$$ **Final answer:** $$\int 6 \sin 2x \cos x \, dx = -\cos 3x - 3 \cos x + C$$