Question

\(\int cot^4x\,dx\) is equal to

• A. \(\frac{cot^3x}{3}-cotx+x+c\)
• B. \(\frac{cot^3x}{3}+cotx+x+c\)
• C. \(-\frac{cot^3x}{3}+cotx+x+c\)
• D. \(\frac{cot^3x}{3}-2cotx+x+c\)

Hint:

Write cot^4 x as cot^2 x (cosec^2 x - 1) and integrate in two parts.

Answer 1

The Correct Option is C

Step 1: Check by differentiation:
Differentiate $-\tfrac13\cot^3x + \cot x + x$.

Step 2: Derivative:
$-\cot^2x\cdot(-\operatorname{cosec}^2x) - \operatorname{cosec}^2x + 1 = \cot^2x\operatorname{cosec}^2x - \cot^2x$ (since $1 - \operatorname{cosec}^2x = -\cot^2x$).

Step 3: Compare:
This equals $\cot^2x(\operatorname{cosec}^2x - 1) = \cot^4x$. So option (C) is the antiderivative.

Final Answer:
The integral is -cot^3 x/3 + cot x + x + c. \[ \boxed{\text{(C) }-\dfrac{\cot^3x}{3}+\cot x+x+c} \]