Question:
If $I_{n}=\int\sin^{n}x dx$, then $I_{6}-\frac{5}{6}I_{4}=$
If $I_{n}=\int\sin^{n}x dx$, then $I_{6}-\frac{5}{6}I_{4}=$
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Memorizing standard reduction patterns like $I_{n} = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2}$ turns long integration-by-parts calculations into single-step algebraic substitutions.
Updated On: Jun 3, 2026
- $\frac{-\sin^{5}x\cos x}{6}$
- $\frac{\sin^{5}x\cos x}{5}$
- $\frac{-\sin^{5}x\cos^{2}x}{6}$
- $\frac{\sin^{5}x\cos^{2}x}{5}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The reduction formula for $I_{n} = \int \sin^{n}x dx$ is derived using integration by parts and is given by the standard identity: $I_{n} = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2}$.
Step 2: Meaning
Rearranging the reduction formula terms helps us isolate the expression required by the question: $I_{n} - \frac{n-1}{n}I_{n-2} = -\frac{\sin^{n-1}x\cos x}{n}$.
Step 3: Analysis
Substitute $n = 6$ directly into this rearranged formula: $I_{6} - \frac{6-1}{6}I_{6-2} = -\frac{\sin^{6-1}x\cos x}{6}$. This simplifies beautifully to: $I_{6} - \frac{5}{6}I_{4} = \frac{-\sin^{5}x\cos x}{6}$.
Step 4: Conclusion
The resulting expression matches option (A) perfectly.
Final Answer: (A)
The reduction formula for $I_{n} = \int \sin^{n}x dx$ is derived using integration by parts and is given by the standard identity: $I_{n} = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2}$.
Step 2: Meaning
Rearranging the reduction formula terms helps us isolate the expression required by the question: $I_{n} - \frac{n-1}{n}I_{n-2} = -\frac{\sin^{n-1}x\cos x}{n}$.
Step 3: Analysis
Substitute $n = 6$ directly into this rearranged formula: $I_{6} - \frac{6-1}{6}I_{6-2} = -\frac{\sin^{6-1}x\cos x}{6}$. This simplifies beautifully to: $I_{6} - \frac{5}{6}I_{4} = \frac{-\sin^{5}x\cos x}{6}$.
Step 4: Conclusion
The resulting expression matches option (A) perfectly.
Final Answer: (A)
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