Integrals of powers of sine and cosine
Problem 4.568 · medium
Evaluate \( \displaystyle \int_0^{\frac{\pi}{4}} \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx \).
- 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).
- \[ \frac{d}{d u} \left(- \frac{u^{3}}{3}\right) = - u^{2} \]∫ -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
- \[ - \frac{\sqrt{2}}{12} + 1 \cdot \frac{1}{3} = \frac{1}{3} - \frac{\sqrt{2}}{12} \]Evaluate from 0.✓ Proved
Answer \( \frac{1}{3} - \frac{\sqrt{2}}{12} \)
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 (error) — Sentence 1 incorrectly describes the substitution: it claims to write the remaining factor as (1−cos²x)^0, which is meaningless, and does not properly justify setting u=cos x. The correct approach is to let u=cos x, du=−sin x dx, and rewrite the integrand as −u² du. The rest of the solution is algebraically correct, but the erroneous sentence misleads the reader about the substitution step.qwen3.6:27b-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)
gpt-oss:20b: fail (error) 2026-10-07 — Sentence 1 incorrectly describes the substitution: it claims to write the remaining factor as (1−cos²x)^0, which is meaningless, and does not properly justify setting u=cos x. The correct approach is to let u=cos x, du=−sin x dx, and rewrite the integrand as −u² du. The rest of the solution is algebraically correct, but the erroneous sentence misleads the reader about the substitution step.qwen3.6:27b-mlx: pass 2026-10-07 — 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: fail (error) 2026-10-07 — The first sentence misstates the substitution: it incorrectly claims to write the remaining factor as (1−cos²x)^0 and does not correctly express cos²x in terms of u. This misleads the reader about the proper substitution.qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies the method for integrating odd powers of sine, performs the substitution and integration accurately, and evaluates the definite integral correctly.
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 2026-10-07 with SymPy 1.14.0.