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: Understanding the Concept:
Use \(\cot^2x = \operatorname{cosec}^2x - 1\) to reduce the power.

Step 2: Split the integrand:
\[ \cot^4x = \cot^2x(\operatorname{cosec}^2x - 1) = \cot^2x\operatorname{cosec}^2x - \cot^2x \]

Step 3: Integrate each part:
First part: let \(t = \cot x\), \(dt = -\operatorname{cosec}^2x\,dx\), so \(\int\cot^2x\operatorname{cosec}^2x\,dx = -\dfrac{\cot^3x}{3}\).
Second part: \(-\int\cot^2x\,dx = -\int(\operatorname{cosec}^2x - 1)dx = \cot x + x\).

Step 4: Combine:
\[ \int\cot^4x\,dx = -\frac{\cot^3x}{3} + \cot x + x + c \]
Options (A), (B) and (D) have the wrong sign on \(\cot^3x/3\) or the wrong coefficient of \(\cot x\).

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