Question:

Step 1: Recall the reduction formula for $I_n$. The given integral is $I_n = \int (\sin x)^n dx$. The standard reduction formula for this integral is: $$I_n = -\frac{1}{n} (\sin x)^{n-1} \cos x + \frac{n-1}{n} I_{n-2}$$ Step 2: Apply the reduction formula for $n=6$. Substitute $n=6$ into the reduction formula to express $I_6$ in terms of $I_4$: $$I_6 = -\frac{1}{6} (\sin x)^{6-1} \cos x + \frac{6-1}{6} I_{6-2}$$ $$I_6 = -\frac{1}{6} (\sin x)^5 \cos x + \frac{5}{6} I_4$$ Step 3: Rearrange the equation to find the desired expression. Multiply the equation from Step 2 by 6 to clear the denominators: $$6 I_6 = -(\sin x)^5 \cos x + 5 I_4$$ Rearrange the terms to find the value of $5I_4 - 6I_6$: $$5 I_4 - 6 I_6 = (\sin x)^5 \cos x$$ Since the integral results in a family of functions, we add the constant of integration, $C$. $$5 I_4 - 6 I_6 = (\sin x)^5 \cos x + C$$ Step 4: Compare the result with the given options. The obtained expression is $5I_4 - 6I_6 = (\sin x)^5 \cos x + C$. Comparing this with the given options: Option 1: (A) $\sin x (\cos x)^5 + C$ Option 2: (B) $\cos x (\sin x)^5 + C$ Option 3: (C) $\frac{\sin 2x}{8}[\cos^2 2x + 1 - 2\cos 2x] + C$ Option 4: (D) $\frac{\sin 2x}{8}[\cos^2 2x + 1 + 2\cos 2x] + C$ The derived expression matches Option 2. Note: Further manipulation of trigonometric identities shows that options (B) and (C) are equivalent, as stated in the original solution. The final answer is $\boxed{(\text{B}) \cos x(\sin x)^5 + C}$.