qingy1337 · July 24, 2025 08:34 · Shawn-the-Sheep · Jul 24, 2025 · qingy1337 · Jul 24, 2025
diff --git a/gistfile1.txt b/gistfile1.txt
 --- Me ---
 What is the integral of cot^4(x) + 3x^2 dx ?

 --- Model ---
 The core task is to compute an indefinite integral involving a trigonometric function and a polynomial. The approach requires applying integration techniques (e.g., substitution, partial fractions) and algebraic manipulation. This demands careful step-by-step analysis, so deep mathematical reasoning is essential here.
 </judge>

 <think_on>
 <think>
 We are to find the integral of cot^4(x) + 3x^2 dx.
 We can split the integral into two parts:
   ∫ cot^4(x) dx + ∫ 3x^2 dx

 The second integral is straightforward:
   ∫ 3x^2 dx = 3 * (x^3/3) = x^3

 For the first integral, we need to integrate cot^4(x). We can use a trigonometric identity to rewrite cot^4(x):
   cot^4(x) = (cot^2(x))^2 = (csc^2(x) - 1)^2   [since cot^2(x) = csc^2(x) - 1]

 So, expand the square:
   (csc^2(x) - 1)^2 = csc^4(x) - 2*csc^2(x) + 1

 Therefore:
   ∫ cot^4(x) dx = ∫ (csc^4(x) - 2*csc^2(x) + 1) dx

 We can split this into:
   ∫ csc^4(x) dx - 2 ∫ csc^2(x) dx + ∫ 1 dx

 We know that the integral of csc^2(x) is -cot(x), and the integral of 1 is x.

 Now, how about ∫ csc^4(x) dx?
 We can write csc^4(x) as csc^2(x) * csc^2(x) = (csc^2(x)) * (1 + cot^2(x))? Actually, note that:
   csc^2(x) = 1 + cot^2(x)

 But that doesn't directly help. Alternatively, we can write:
   ∫ csc^4(x) dx = ∫ csc^2(x) * csc^2(x) dx = ∫ (1 + cot^2(x)) * csc^2(x) dx

 Now, we can use substitution. Let u = cot(x), then du = -csc^2(x) dx, so csc^2(x) dx = -du.

 Then:
   ∫ (1 + u^2) * (-du) = -∫ (1 + u^2) du = - [ u + (u^3)/3 ] + C = -u - u^3/3 + C

 Substitute back:
   -cot(x) - (cot^3(x))/3 + C

 Therefore, putting together:

   ∫ cot^4(x) dx = [ -cot(x) - (cot^3(x))/3 ] - 2 * (-cot(x)) + x + C

 Simplify:
   = -cot(x) - (cot^3(x))/3 + 2*cot(x) + x + C
   = ( -cot(x) + 2*cot(x) ) - (cot^3(x))/3 + x + C
   = cot(x) - (cot^3(x))/3 + x + C

 Alternatively, we can factor:
   = cot(x) * (1 - cot^2(x)/3) + x + C

 But note: 1 - cot^2(x)/3 = (3 - cot^2(x))/3, so:
   = cot(x) * (3 - cot^2(x)) / 3 + x + C

 However, we can also leave it as:
   = x + cot(x) - (1/3) cot^3(x) + C

 Therefore, the entire integral is:

   ∫ [cot^4(x) + 3x^2] dx = [x + cot(x) - (1/3) cot^3(x)] + x^3 + C

 But wait, note that we have two x terms? Actually, the integral of 1 dx in the cot^4 part gave an x, and the integral of 3x^2 gave x^3.

 So combining:

   = x^3 + x + cot(x) - (1/3) cot^3(x) + C

 However, let me double-check the steps.

 Alternatively, we can write the entire integral without splitting the constant of integration until the end.

 Final answer:

   ∫ (cot^4(x) + 3x^2) dx = x^3 + x + cot(x) - (1/3) cot^3(x) + C

 But note: we can also write the cotangent terms as:

   cot(x) - (1/3) cot^3(x) = cot(x) (1 - (1/3) cot^2(x))

 However, the expanded form is acceptable.

 Let me verify by differentiating:

   d/dx [ x^3 ] = 3x^2
   d/dx [ x ] = 1
   d/dx [ cot(x) ] = -csc^2(x)
   d/dx [ - (1/3) cot^3(x) ] = - (1/3) * 3 cot^2(x) * (-csc^2(x)) = cot^2(x) * csc^2(x)

 So the derivative of the entire expression is:

   3x^2 + 1 - csc^2(x) + cot^2(x) * csc^2(x)

 But note that cot^2(x) * csc^2(x) = (cos^2(x)/sin^2(x)) * (1/sin^2(x)) = cos^2(x)/sin^4(x)

 And we know that the original function is cot^4(x) + 3x^2.

 We need to check if:

   3x^2 + 1 - csc^2(x) + cot^2(x)*csc^2(x) = cot^4(x) + 3x^2

 Subtract 3x^2 from both sides:

   1 - csc^2(x) + cot^2(x)*csc^2(x) = cot^4(x)

 Now, note that:

   cot^4(x) = (cot^2(x))^2 = (csc^2(x) - 1)^2 = csc^4(x) - 2*csc^2(x) + 1

 So we have:

   Right-hand side: cot^4(x) = csc^4(x) - 2*csc^2(x) + 1

   Left-hand side: 1 - csc^2(x) + cot^2(x)*csc^2(x)

 But note: cot^2(x)*csc^2(x) = (csc^2(x) - 1) * csc^2(x) = csc^4(x) - csc^2(x)

 Therefore, left-hand side:

   1 - csc^2(x) + (csc^4(x) - csc^2(x)) = 1 - 2*csc^2(x) + csc^4(x)

 Which is exactly the same as the right-hand side.

 Hence, verified.

 Therefore, the integral is:

   x^3 + x + cot(x) - (1/3) cot^3(x) + C

 Alternatively, we can write:

   x^3 + x + cot(x) (1 - (1/3) cot^2(x)) + C

 But the first form is standard.

 So the final answer is:

   ∫ (cot^4(x) + 3x^2) dx = x^3 + x + cot(x) - \frac{1}{3} \cot^3(x) + C
 </think>

 <answer>
 The integral of \(\cot^4(x) + 3x^2  dx\) is computed as follows:

 ### Step 1: Split the integral
 \[
 \int (\cot^4(x) + 3x^2)  dx = \int \cot^4(x)  dx + \int 3x^2  dx
 \]

 ### Step 2: Integrate \(3x^2\)
 \[
 \int 3x^2  dx = 3 \cdot \frac{x^3}{3} = x^3
 \]

 ### Step 3: Integrate \(\cot^4(x)\)
 Use the identity \(\cot^2(x) = \csc^2(x) - 1\):
 \[
 \cot^4(x) = (\cot^2(x))^2 = (\csc^2(x) - 1)^2 = \csc^4(x) - 2\csc^2(x) + 1
 \]
 So,
 \[
 \int \cot^4(x)  dx = \int (\csc^4(x) - 2\csc^2(x) + 1)  dx = \int \csc^4(x)  dx - 2 \int \csc^2(x)  dx + \int 1  dx
 \]

 - **Term 1:** \(\int \csc^2(x)  dx = -\cot(x)\)
 - **Term 2:** \(\int 1  dx = x\)
 - **Term 3:** \(\int \csc^4(x)  dx\)
  Rewrite \(\csc^4(x) = \csc^2(x) \cdot \csc^2(x) = (1 + \cot^2(x)) \csc^2(x)\). Use substitution \(u = \cot(x)\), \(du = -\csc^2(x)  dx\):
  \[
  \int \csc^4(x)  dx = \int (1 + u^2) (-du) = -\int (1 + u^2)  du = -(u + \frac{u^3}{3}) + C = -\cot(x) - \frac{\cot^3(x)}{3}
  \]

 Combine all terms for \(\int \cot^4(x)  dx\):
 \[
 \left(-\cot(x) - \frac{\cot^3(x)}{3}\right) - 2(-\cot(x)) + x = -\cot(x) - \frac{\cot^3(x)}{3} + 2\cot(x) + x = x + \cot(x) - \frac{\cot^3(x)}{3}
 \]

 ### Step 4: Combine results
 \[
 \int (\cot^4(x) + 3x^2)  dx = \left(x + \cot(x) - \frac{\cot^3(x)}{3}\right) + x^3 + C
 \]
 where \(C\) is the constant of integration.

 ### Final Answer
 \[
 \boxed{x^{3} + x + \cot\left(x\right) - \dfrac{\cot^{3}\left(x\right)}{3} + C}
 \]
 </answer>
	--- Me ---
	What is the integral of cot^4(x) + 3x^2 dx ?

	--- Model ---
	The core task is to compute an indefinite integral involving a trigonometric function and a polynomial. The approach requires applying integration techniques (e.g., substitution, partial fractions) and algebraic manipulation. This demands careful step-by-step analysis, so deep mathematical reasoning is essential here.
	</judge>

	<think_on>
	<think>
	We are to find the integral of cot^4(x) + 3x^2 dx.
	We can split the integral into two parts:
	∫ cot^4(x) dx + ∫ 3x^2 dx

	The second integral is straightforward:
	∫ 3x^2 dx = 3 * (x^3/3) = x^3

	For the first integral, we need to integrate cot^4(x). We can use a trigonometric identity to rewrite cot^4(x):
	cot^4(x) = (cot^2(x))^2 = (csc^2(x) - 1)^2 [since cot^2(x) = csc^2(x) - 1]

	So, expand the square:
	(csc^2(x) - 1)^2 = csc^4(x) - 2*csc^2(x) + 1

	Therefore:
	∫ cot^4(x) dx = ∫ (csc^4(x) - 2*csc^2(x) + 1) dx

	We can split this into:
	∫ csc^4(x) dx - 2 ∫ csc^2(x) dx + ∫ 1 dx

	We know that the integral of csc^2(x) is -cot(x), and the integral of 1 is x.

	Now, how about ∫ csc^4(x) dx?
	We can write csc^4(x) as csc^2(x) * csc^2(x) = (csc^2(x)) * (1 + cot^2(x))? Actually, note that:
	csc^2(x) = 1 + cot^2(x)

	But that doesn't directly help. Alternatively, we can write:
	∫ csc^4(x) dx = ∫ csc^2(x) * csc^2(x) dx = ∫ (1 + cot^2(x)) * csc^2(x) dx

	Now, we can use substitution. Let u = cot(x), then du = -csc^2(x) dx, so csc^2(x) dx = -du.

	Then:
	∫ (1 + u^2) * (-du) = -∫ (1 + u^2) du = - [ u + (u^3)/3 ] + C = -u - u^3/3 + C

	Substitute back:
	-cot(x) - (cot^3(x))/3 + C

	Therefore, putting together:

	∫ cot^4(x) dx = [ -cot(x) - (cot^3(x))/3 ] - 2 * (-cot(x)) + x + C

	Simplify:
	= -cot(x) - (cot^3(x))/3 + 2*cot(x) + x + C
	= ( -cot(x) + 2*cot(x) ) - (cot^3(x))/3 + x + C
	= cot(x) - (cot^3(x))/3 + x + C

	Alternatively, we can factor:
	= cot(x) * (1 - cot^2(x)/3) + x + C

	But note: 1 - cot^2(x)/3 = (3 - cot^2(x))/3, so:
	= cot(x) * (3 - cot^2(x)) / 3 + x + C

	However, we can also leave it as:
	= x + cot(x) - (1/3) cot^3(x) + C

	Therefore, the entire integral is:

	∫ [cot^4(x) + 3x^2] dx = [x + cot(x) - (1/3) cot^3(x)] + x^3 + C

	But wait, note that we have two x terms? Actually, the integral of 1 dx in the cot^4 part gave an x, and the integral of 3x^2 gave x^3.

	So combining:

	= x^3 + x + cot(x) - (1/3) cot^3(x) + C

	However, let me double-check the steps.

	Alternatively, we can write the entire integral without splitting the constant of integration until the end.

	Final answer:

	∫ (cot^4(x) + 3x^2) dx = x^3 + x + cot(x) - (1/3) cot^3(x) + C

	But note: we can also write the cotangent terms as:

	cot(x) - (1/3) cot^3(x) = cot(x) (1 - (1/3) cot^2(x))

	However, the expanded form is acceptable.

	Let me verify by differentiating:

	d/dx [ x^3 ] = 3x^2
	d/dx [ x ] = 1
	d/dx [ cot(x) ] = -csc^2(x)
	d/dx [ - (1/3) cot^3(x) ] = - (1/3) * 3 cot^2(x) * (-csc^2(x)) = cot^2(x) * csc^2(x)

	So the derivative of the entire expression is:

	3x^2 + 1 - csc^2(x) + cot^2(x) * csc^2(x)

	But note that cot^2(x) * csc^2(x) = (cos^2(x)/sin^2(x)) * (1/sin^2(x)) = cos^2(x)/sin^4(x)

	And we know that the original function is cot^4(x) + 3x^2.

	We need to check if:

	3x^2 + 1 - csc^2(x) + cot^2(x)*csc^2(x) = cot^4(x) + 3x^2

	Subtract 3x^2 from both sides:

	1 - csc^2(x) + cot^2(x)*csc^2(x) = cot^4(x)

	Now, note that:

	cot^4(x) = (cot^2(x))^2 = (csc^2(x) - 1)^2 = csc^4(x) - 2*csc^2(x) + 1

	So we have:

	Right-hand side: cot^4(x) = csc^4(x) - 2*csc^2(x) + 1

	Left-hand side: 1 - csc^2(x) + cot^2(x)*csc^2(x)

	But note: cot^2(x)csc^2(x) = (csc^2(x) - 1) csc^2(x) = csc^4(x) - csc^2(x)

	Therefore, left-hand side:

	1 - csc^2(x) + (csc^4(x) - csc^2(x)) = 1 - 2*csc^2(x) + csc^4(x)

	Which is exactly the same as the right-hand side.

	Hence, verified.

	Therefore, the integral is:

	x^3 + x + cot(x) - (1/3) cot^3(x) + C

	Alternatively, we can write:

	x^3 + x + cot(x) (1 - (1/3) cot^2(x)) + C

	But the first form is standard.

	So the final answer is:

	∫ (cot^4(x) + 3x^2) dx = x^3 + x + cot(x) - \frac{1}{3} \cot^3(x) + C
	</think>

	<answer>
	The integral of \(\cot^4(x) + 3x^2 dx\) is computed as follows:

	### Step 1: Split the integral
	\[
	\int (\cot^4(x) + 3x^2) dx = \int \cot^4(x) dx + \int 3x^2 dx
	\]

	### Step 2: Integrate \(3x^2\)
	\[
	\int 3x^2 dx = 3 \cdot \frac{x^3}{3} = x^3
	\]

	### Step 3: Integrate \(\cot^4(x)\)
	Use the identity \(\cot^2(x) = \csc^2(x) - 1\):
	\[
	\cot^4(x) = (\cot^2(x))^2 = (\csc^2(x) - 1)^2 = \csc^4(x) - 2\csc^2(x) + 1
	\]
	So,
	\[
	\int \cot^4(x) dx = \int (\csc^4(x) - 2\csc^2(x) + 1) dx = \int \csc^4(x) dx - 2 \int \csc^2(x) dx + \int 1 dx
	\]

	- Term 1: \(\int \csc^2(x) dx = -\cot(x)\)
	- Term 2: \(\int 1 dx = x\)
	- Term 3: \(\int \csc^4(x) dx\)
	Rewrite \(\csc^4(x) = \csc^2(x) \cdot \csc^2(x) = (1 + \cot^2(x)) \csc^2(x)\). Use substitution \(u = \cot(x)\), \(du = -\csc^2(x) dx\):
	\[
	\int \csc^4(x) dx = \int (1 + u^2) (-du) = -\int (1 + u^2) du = -(u + \frac{u^3}{3}) + C = -\cot(x) - \frac{\cot^3(x)}{3}
	\]

	Combine all terms for \(\int \cot^4(x) dx\):
	\[
	\left(-\cot(x) - \frac{\cot^3(x)}{3}\right) - 2(-\cot(x)) + x = -\cot(x) - \frac{\cot^3(x)}{3} + 2\cot(x) + x = x + \cot(x) - \frac{\cot^3(x)}{3}
	\]

	### Step 4: Combine results
	\[
	\int (\cot^4(x) + 3x^2) dx = \left(x + \cot(x) - \frac{\cot^3(x)}{3}\right) + x^3 + C
	\]
	where \(C\) is the constant of integration.

	### Final Answer
	\[
	\boxed{x^{3} + x + \cot\left(x\right) - \dfrac{\cot^{3}\left(x\right)}{3} + C}
	\]
	</answer>
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