Integral of \( \displaystyle \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} \)
Problem 4.463 · medium
Find \( \displaystyle \int \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\, dx \]integralStart with the integral of the given function.✓ Proved
- \[ = \int \cot{\left(2 x \right)}\, dx \]rewriteRewrite the quotient using the cotangent function.✓ Proved
- \[ = \int \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\, dx \]substitutionLet u = sin(2*x), then du = 2*cos(2*x)dx.✓ Proved
- \[ = 0.5 \ln{\left(\sin{\left(2 x \right)} \right)} \]antiderivativeIntegrate the resulting expression.✓ Proved
Answer \( \frac{\ln{\left(\sin{\left(2 x \right)} \right)}}{2} + C \)
Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where sin(2*x) = 0 cot has poles at multiples of pi |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 cot has poles at multiples of pi undefined where sin(2*x) = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where sin(2*x) = 0 log is undefined for non-positive arguments |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated the stated antiderivative back to the integrand |
Reviewers
gpt-oss:20b: fail (error) — Step 3 claims a substitution but leaves the integrand unchanged; it also omits the factor 1/2 that arises from du=2cos(2x)dx. Step 4 gives 0.5*log(sin(2*x)) without the absolute value required for the logarithm of a real argument.qwen3.6:27b-mlx: fail (error) — Step 3 claims to use substitution but fails to show the intermediate step of rewriting the integral in terms of u (i.e., Integral(1/u, u) * 1/2). It jumps directly from the original integral to the final answer, violating the one-change-per-step rule and skipping the algebraic substitution logic.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-06 — Step 3 claims a substitution but leaves the integrand unchanged; it also omits the factor 1/2 that arises from du=2cos(2x)dx. Step 4 gives 0.5*log(sin(2*x)) without the absolute value required for the logarithm of a real argument.qwen3.6:27b-mlx: fail (error) 2026-10-06 — Step 3 claims to use substitution but fails to show the intermediate step of rewriting the integral in terms of u (i.e., Integral(1/u, u) * 1/2). It jumps directly from the original integral to the final answer, violating the one-change-per-step rule and skipping the algebraic substitution logic.qwen3.6:27b-mlx: fail (error) 2026-10-06 — Step 3 claims to use substitution but leaves the expression in terms of x (cos(2*x)/sin(2*x)) instead of rewriting it in terms of u. Step 4 jumps from the un-substituted integral to the final answer without showing the integration of 1/u or the back-substitution, violating the one-change-per-step rule.gpt-oss:20b: fail (error) 2026-10-06 — Step 3 incorrectly repeats the original integrand instead of applying the substitution; it should transform the integral into ∫(1/u)(du/2). Also step 4 omits the absolute value in the logarithm, which is required for the antiderivative of 1/u.
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
gemma4:26b, checked 2026-10-06 with SymPy 1.14.0.