Integrals giving inverse trig functions
Problem 4.633 · medium
Evaluate \( \displaystyle \int \frac{1}{\sqrt{- x^{2} + 4 x + 21}}\, dx \).
- \[ - x^{2} + 4 x + 21 = 25 - \left(x - 2\right)^{2} \]Complete the square.✓ Proved
- With u = x − 2 and a = 5, this is c·∫ du/√(a² − u²)Reviewed
- \[ \frac{d}{d x} \operatorname{asin}{\left(\frac{x}{5} - \frac{2}{5} \right)} = \frac{1}{\sqrt{- x^{2} + 4 x + 21}} \]An antiderivative is asin(x/5 - 2/5); differentiate to confirm.✓ Proved
Answer \( \operatorname{asin}{\left(\frac{x}{5} - \frac{2}{5} \right)} + C \)
Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | difference quotient of the answer at x = -2.9 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07gpt-oss:20b: fail (misleading) 2026-10-07 — The sentence claims the integral becomes "c·∫ du/√(a²−u²)" but there is no extra constant factor; the correct form is ∫ du/√(25−u²). This misleads a student about the presence of an unnecessary constant multiplier.qwen3.6:27b-mlx: pass 2026-10-07
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
generator:structured/inverse_trig_integral, checked 2026-10-07 with SymPy 1.14.0.