Integrals of powers of sine and cosine
Problem 4.856 · medium
Evaluate \( \displaystyle \int_0^{\frac{\pi}{4}} \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx \).
- 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
- \[ - \frac{\sqrt{2}}{40} + 1 \cdot \frac{1}{5} = \frac{1}{5} - \frac{\sqrt{2}}{40} \]Evaluate from 0.✓ Proved
Answer \( \frac{1}{5} - \frac{\sqrt{2}}{40} \)
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: inconclusive — reviewer response could not be parsed: {"verdict":"fail","severity":"misleading","notes":"The first sentence incorrectly states that the remaining factor is $(1-\cos^2x)^0$, which is just 1 and ignores the $\cos^4x$ term. It also omits theqwen3.6:27b-mlx: pass — The solution correctly identifies the substitution method for an odd power of sine, performs the integration and differentiation checks accurately, and evaluates the definite integral correctly.
Every verdict on record (4)
gpt-oss:20b: inconclusive 2026-10-10 — reviewer response could not be parsed: {"verdict":"fail","severity":"misleading","notes":"The first sentence incorrectly states that the remaining factor is $(1-\cos^2x)^0$, which is just 1 and ignores the $\cos^4x$ term. It also omits theqwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the substitution method for an odd power of sine, performs the integration and differentiation checks accurately, and evaluates the definite integral correctly.gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the substitution method for an odd power of sine, performs the integration and evaluation accurately, and arrives at the correct final answer.
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 2026-10-10 with SymPy 1.14.0.