Question

Assertion (A): If \(I_n=\int \cot^n x \, dx\), then \(I_6+I_4=\dfrac{-\cot^5 x}{5}\)
Reason (R): \[ \int \cot^n x \, dx=\frac{-\cot^{\,n-1}x}{n-1}-\int \cot^{\,n-2}x \, dx \]

• A. A is false, R is false
• B. A is true, R is true
• C. A is true, R is false
• D. A is false, R is true

Hint:

Remember the reduction formula: \[ I_n=\int \cot^n x\,dx = -\frac{\cot^{\,n-1}x}{n-1}-I_{n-2} \] This formula is frequently used in integration problems involving powers of trigonometric functions.

Answer 1

The Correct Option is C

Step 1: Verify the assertion.
Given, \[ I_n=\int \cot^n x \, dx \] Using the standard reduction formula for \(\cot^n x\), \[ I_n=-\frac{\cot^{\,n-1}x}{n-1}-I_{n-2} \] Substituting \(n=6\), \[ I_6=-\frac{\cot^5 x}{5}-I_4 \] Rearranging, \[ I_6+I_4=-\frac{\cot^5 x}{5} \] Hence, Assertion (A) is true.

Step 2: Verify the reason.
The given reason states \[ \int \cot^n x \, dx = \frac{-\cot^{\,n-1}x}{n} -\int \cot^{\,n-2}x \, dx \] But the correct reduction formula is \[ \int \cot^n x \, dx = -\frac{\cot^{\,n-1}x}{n-1} -\int \cot^{\,n-2}x \, dx \] The denominator should be \(n-1\), not \(n\).
Therefore, the given reason is false.

Step 3: Final Conclusion.
Assertion (A) is true, but Reason (R) is false.
Hence, the correct option is \[ \boxed{\text{(3) A is true, R is false}} \]