Integrals of powers of sine and cosine
Problem 4.713 · medium
Evaluate \( \displaystyle \int_0^{\frac{\pi}{2}} \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx \).
- 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).
- \[ \frac{d}{d u} \left(- \frac{u^{5}}{5}\right) = - u^{4} \]∫ -u**4 du = -u**5/5.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{\cos^{5}{\left(x \right)}}{5}\right) = \sin{\left(x \right)} \cos^{4}{\left(x \right)} \]Differentiating the answer returns the integrand.✓ Proved
- \[ 1 \cdot \frac{1}{5} = \frac{1}{5} \]Evaluate from 0.✓ Proved
Answer \( \frac{1}{5} \)
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 sentence claims to rewrite the remaining factor as (1−cos²x)^0, which is incorrect and confusing; the correct step is simply cos^4x = u^4. This misleads a student about how to handle the integrand.qwen3.6:27b-mlx: pass — The solution correctly identifies the substitution method for an odd power of sine, performs the integration and differentiation checks correctly, and evaluates the definite integral accurately.
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-08 — The sentence claims to rewrite the remaining factor as (1−cos²x)^0, which is incorrect and confusing; the correct step is simply cos^4x = u^4. This misleads a student about how to handle the integrand.qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly identifies the substitution method for an odd power of sine, performs the integration and differentiation checks correctly, and evaluates the definite integral accurately.gpt-oss:20b: fail (error) 2026-10-08 — Sentence 1 incorrectly states that the remaining factor is (1−cos²x)^0, which is 1 and does not reflect the actual integrand. The correct approach is to set u=cos x and use the entire cos⁴x as u⁴. The rest of the solution is fine.qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly identifies the substitution method for an odd power of sine, performs the integration and back-substitution 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 2026-10-08 with SymPy 1.14.0.