Integrals of powers of sine and cosine
Problem 4.708 · medium
Evaluate \( \displaystyle \int_0^{\frac{\pi}{4}} \cos^{3}{\left(x \right)}\, dx \).
- The power of cosine is odd: keep one cos x for du, write the rest as (1 − sin²x)^1, and let u = sin x.
- \[ \frac{d}{d u} \left(- \frac{u^{3}}{3} + u\right) = 1 - u^{2} \]∫ 1 - u**2 du = -u**3/3 + u.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) = \cos^{3}{\left(x \right)} \]Differentiating the answer returns the integrand.✓ Proved
- \[ \frac{5 \sqrt{2}}{12} \]Evaluate from 0.✓ Proved
Answer \( \frac{5 \sqrt{2}}{12} \)
Lines: 3 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 |
| 4 | ✓ 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 | numerical quadrature |
Reviewers
gpt-oss:20b: fail (misleading) — The solution never evaluates the antiderivative at the limits 0 and π/4, yet claims the result is 5√2/12. This misleads a student into thinking the answer is justified without showing the necessary computation.qwen3.6:27b-mlx: pass — The solution correctly identifies the method for integrating an odd power of cosine, verifies the antiderivative, and correctly evaluates the definite integral.
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-08 — The solution never evaluates the antiderivative at the limits 0 and π/4, yet claims the result is 5√2/12. This misleads a student into thinking the answer is justified without showing the necessary computation.qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly identifies the method for integrating an odd power of cosine, verifies the antiderivative, and correctly evaluates the definite integral.gpt-oss:20b: fail (misleading) 2026-10-08 — Sentence 4 claims to evaluate the integral but merely repeats the answer without showing the calculation; it does not demonstrate that the bounds were applied correctly, which could mislead a student.qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies the standard technique for integrating odd powers of cosine and verifies the result via differentiation and evaluation.
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 2026-10-08 with SymPy 1.14.0.