Integrals of powers of sine and cosine
Problem 4.417 · medium
Evaluate \( \displaystyle \int_0^{\pi} \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx \).
- The power of sine is odd: keep one sin x for du, write the rest as (1 − cos²x)^0, and let u = cos x (du = −sin x dx).Reviewed
- \[ \frac{d}{d u} \left(- \frac{u^{3}}{3}\right) = - u^{2} \]∫ -u**2 du = -u**3/3.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{\cos^{3}{\left(x \right)}}{3}\right) = \sin{\left(x \right)} \cos^{2}{\left(x \right)} \]Differentiating the answer returns the integrand.✓ Proved
- \[ \frac{2}{3} \]Evaluate from 0.✓ Proved
Answer \( \frac{2}{3} \)
Lines: 3 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly identifies the substitution method for an odd power of sine, verifies the antiderivative, and correctly evaluates the definite integral.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the substitution method for an odd power of sine, verifies the antiderivative, and correctly evaluates the definite integral.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims the power of sine is odd to justify the substitution method, but the power of sine is 1, which is odd, so that part is technically correct, however, the explanation 'write the rest as (1 - cos^2x)^0' is confusing and unnecessary for sin^1(x). More critically, the final evaluation step '2/3 = 2/3' is presented as a check rather than a derivation, and the intermediate steps do not show the definite integral evaluation clearly. The main issue is that step 1 describes a method for odd powers of sine generally, but applies it to sin^1(x) in a way that suggests a more complex decomposition than needed, and the final line is not a proper evaluation step but a tautology check. However, the biggest error is likely in the interpretation of the steps: Step 2 and 3 are indefinite integrals/checks. Step 4 just states the answer. The solution lacks the actual evaluation of the definite integral using the bounds. It jumps from finding the antiderivative to stating the final value without showing the substitution of bounds [-cos^3(x)/3]_0^pi = -(-1)^3/3 - (-1)^3/3 = 1/3 + 1/3 = 2/3. The line '2/3 = 2/3' is not a valid mathematical step in a solution, it's a verification. A solution must show the calculation.gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.