Integrals of powers of sine and cosine
Problem 4.859 · medium
Evaluate \( \displaystyle \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx \).
- The power of cosine is odd: keep one cos x for du, write the rest as (1 − sin²x)^0, and let u = sin x.
- \[ \frac{d}{d u} \frac{u^{3}}{3} = u^{2} \]∫ u**2 du = u**3/3.✓ Proved
- \[ \frac{d}{d x} \frac{\sin^{3}{\left(x \right)}}{3} = \sin^{2}{\left(x \right)} \cos{\left(x \right)} \]Differentiating the answer returns the integrand.✓ Proved
Answer \( \frac{\sin^{3}{\left(x \right)}}{3} + C \)
Lines: 2 proved, 1 not checked. 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 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 quotients of the answer match the integrand |
Reviewers
gpt-oss:20b: fail (error) — Sentence 1 incorrectly rewrites the integrand: it claims to replace the remaining σ^2x with (1−σ^2x)^0, which is 1, thereby changing the integrand. The correct approach is to set u=sin x and keep the sin^2x factor; no such rewrite is needed. This misstatement could mislead a student.qwen3.6:27b-mlx: pass — The solution correctly identifies the substitution method for an integrand with an odd power of cosine. The steps are logically sound and the final answer is verified.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-10 — Sentence 1 incorrectly rewrites the integrand: it claims to replace the remaining σ^2x with (1−σ^2x)^0, which is 1, thereby changing the integrand. The correct approach is to set u=sin x and keep the sin^2x factor; no such rewrite is needed. This misstatement could mislead a student.qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the substitution method for an integrand with an odd power of cosine. The steps are logically sound and the final answer is verified.gpt-oss:20b: fail (error) 2026-10-10 — The first sentence incorrectly rewrites the integrand: it claims to write the remaining factor as (1−sin²x)^0, which equals 1, but the correct substitution is sin²x = u². This misstates the setup and would mislead a student.qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the substitution method for an odd power of cosine and verifies the result by differentiation.
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/trig_powers_integral, checked 2026-10-10 with SymPy 1.14.0.