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: Recall the cotangent reduction.
The correct reduction formula is $\displaystyle\int\cot^n x\,dx=-\dfrac{\cot^{n-1}x}{n-1}-\int\cot^{n-2}x\,dx$, so $I_n=-\dfrac{\cot^{n-1}x}{n-1}-I_{n-2}$.
Step 2: Test the assertion with $n=6$.
Putting $n=6$, $I_6=-\dfrac{\cot^5 x}{5}-I_4$, hence $I_6+I_4=-\dfrac{\cot^5 x}{5}$. So Assertion A is true.
Step 3: Read the Reason carefully.
The Reason writes the formula with denominator $n$, that is $\int\cot^n x\,dx=\dfrac{-\cot^{n-1}x}{n}-\int\cot^{n-2}x\,dx$.
Step 4: Compare with the true formula.
The genuine denominator is $n-1$, not $n$. So the Reason as stated is false.
Step 5: Decide A and R together.
A is true and R is false.
Step 6: Choose the option.
This matches option 3, A is true, R is false.
\[ \boxed{\text{A is true, R is false}} \]