Integrals of powers of sine and cosine
Problem 4.1017 · medium
Evaluate \( \displaystyle \int_0^{\pi} \cos^{3}{\left(x \right)}\, dx \).
- 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
- \[ 0 \]Evaluate from 0.✓ Proved
Answer \( 0 \)
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: passqwen3.6:27b-mlx: fail (error) — The solution fails to evaluate the antiderivative at the upper limit x = π. Since sin(π) = 0, the term is 0, but the step explicitly only shows the evaluation at the lower limit (0), omitting the necessary subtraction of the upper limit value in the Fundamental Theorem of Calculus application.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: fail (error) 2026-10-13 — The solution fails to evaluate the antiderivative at the upper limit x = π. Since sin(π) = 0, the term is 0, but the step explicitly only shows the evaluation at the lower limit (0), omitting the necessary subtraction of the upper limit value in the Fundamental Theorem of Calculus application.gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: fail (error) 2026-10-13 — The solution fails to evaluate the antiderivative at the upper limit x = π. Since sin(π) = 0, the term at the upper limit is also 0, but the step explicitly labeled 'Evaluate from 0' ignores the upper bound entirely, making the logic incomplete and technically incorrect as a definite integral 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 2026-10-13 with SymPy 1.14.0.