Integral of \( \displaystyle \tan^{2}{\left(x - 1 \right)} \)
Problem 4.178 · easy
Find \( \displaystyle \int \tan^{2}{\left(x - 1 \right)} \, dx \). (Omit the constant of integration.)
- \[ \int \tan^{2}{\left(x - 1 \right)}\, dx \]integralStart with the integral of the given function.✓ Proved
- \[ = \int \left(\sec^{2}{\left(x - 1 \right)} - 1\right)\, dx \]trig-identityUse the identity tan(u)**2 = sec(u)**2 - 1.Reviewed
- \[ = - \int 1\, dx + \int \sec^{2}{\left(x - 1 \right)}\, dx \]linearitySplit the integral into two parts.✓ Proved
- \[ = - x + \int \sec^{2}{\left(x - 1 \right)}\, dx \]antiderivativeEvaluate the integral of the constant 1.✓ Proved
- \[ = - x + \tan{\left(x - 1 \right)} \]antiderivativeEvaluate the integral of sec(x - 1)**2.Reviewed
Answer \( - x + \tan{\left(x - 1 \right)} + C \)
Lines: 4 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 tan(x - 1) - Integral(sec(x - 1)**2, x); no point in the sample was defined on both lines tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 sec has poles at odd multiples of pi/2 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 sec has poles at odd multiples of pi/2 |
| 5 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | simplify left -tan(x - 1) + Integral(sec(x - 1)**2, x); no point in the sample was defined on both lines sec has poles at odd multiples of pi/2 tan 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 tan has poles at odd multiples of pi/2 |
| 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 trigonometric identities and linearity of integration. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly applies trigonometric identities and linearity of integration. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.gpt-oss:20b: pass 2026-09-29qwen3.6:27b-mlx: pass 2026-09-29gpt-oss:20b: pass 2026-09-29
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 2026-09-29 with SymPy 1.14.0.