Question:

$\int \cot^2 x \, dx =$

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A common mistake is choosing Option (A) by forgetting that the integral of $-1$ is $-x$. Similarly, for $\tan^2 x$, the integral is $\tan x - x + c$. Notice the symmetry in how these squared trig functions integrate.
  • $-\cot x + x + c$
  • $\cot x - x + c$
  • $\cot^2 x - x + c$
  • $-\cot x - x + c$
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The Correct Option is D

Solution and Explanation

To integrate $\cot^2 x$, we must first transform it into a form that has a direct standard integral. Step 1: Use Trigonometric Identity Recall the Pythagorean identity for cotangent: $$1 + \cot^2 x = \text{cosec}^2 x$$ Rearranging to solve for $\cot^2 x$: $$\cot^2 x = \text{cosec}^2 x - 1$$

Step 2: Rewrite the Integral Substitute the identity into the original integral: $$\int \cot^2 x \, dx = \int (\text{cosec}^2 x - 1) \, dx$$

Step 3: Integrate term by term The integral can be split: $$\int \text{cosec}^2 x \, dx - \int 1 \, dx$$ From standard integral formulas: $\int \text{cosec}^2 x \, dx = -\cot x$ $\int 1 \, dx = x$

Step 4: Final Answer Combining the parts and adding the constant of integration ($c$): $$\text{Integral} = -\cot x - x + c$$
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