Integral of \( \displaystyle \frac{\sin{\left(3 x - 1 \right)}}{\cos^{2}{\left(3 x - 1 \right)}} \)
Problem 4.1000 · medium
Find \( \displaystyle \int \frac{\sin{\left(3 x - 1 \right)}}{\cos^{2}{\left(3 x - 1 \right)}} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{\sin{\left(3 x - 1 \right)}}{\cos^{2}{\left(3 x - 1 \right)}}\, dx \]integralStart with the integral of the given function.✓ Proved
- \[ = \int \frac{\tan{\left(3 x - 1 \right)}}{\cos{\left(3 x - 1 \right)}}\, dx \]rewriteRewrite the numerator using the identity sin(u) = tan(u) * cos(u).Reviewed
- \[ = \int \tan{\left(3 x - 1 \right)} \sec{\left(3 x - 1 \right)}\, dx \]rewriteRewrite 1/cos(u) as sec(u).Reviewed
- \[ = \frac{\sec{\left(3 x - 1 \right)}}{3} \]antiderivativeIntegrate the product of tangent and secant using the substitution rule.✓ Proved
Answer \( \frac{1}{3 \cos{\left(3 x - 1 \right)}} + C \)
Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | simplify left -Integral(tan(3*x - 1)/cos(3*x - 1), x) + 1/(3*cos(3*x - 1)); no point in the sample was defined on both lines undefined where cos(3*x - 1) = 0 tan has poles at odd multiples of pi/2 |
| 3 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | simplify left -sec(3*x - 1)/3 + Integral(tan(3*x - 1)/cos(3*x - 1), x); no point in the sample was defined on both lines tan has poles at odd multiples of pi/2 undefined where cos(3*x - 1) = 0 sec has poles at odd multiples of pi/2 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 undefined where cos(3*x - 1) = 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated the stated antiderivative back to the integrand |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies algebraic rewrites to transform the integrand into a standard form and then applies the known antiderivative rule for tan(u)sec(u) with the appropriate chain rule factor. Each step changes only one aspect of the expression.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly applies algebraic rewrites to transform the integrand into a standard form and then applies the known antiderivative rule for tan(u)sec(u) with the appropriate chain rule factor. Each step changes only one aspect of the expression.gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly applies trigonometric identities to rewrite the integrand into a standard form and then applies the known antiderivative rule for secant times tangent, accounting for the chain rule factor.
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 2026-10-13 with SymPy 1.14.0.