Integrals of powers of sine and cosine
Problem 4.712 · medium
Evaluate \( \displaystyle \int_0^{\pi} \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx \).
- The power of cosine is odd: keep one cos x for du, write the rest as (1 − sin²x)^0, and let u = sin x.
- \[ \frac{d}{d u} \frac{u^{5}}{5} = u^{4} \]∫ u**4 du = u**5/5.✓ Proved
- \[ \frac{d}{d x} \frac{\sin^{5}{\left(x \right)}}{5} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} \]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 = pi. Since sin(pi) = 0, the term at the upper limit is also 0, but omitting this step makes the calculation incomplete and logically insufficient to justify the result.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to evaluate the antiderivative at the upper limit x = pi. Since sin(pi) = 0, the term at the upper limit is also 0, but omitting this step makes the calculation incomplete and logically insufficient to justify the result.gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to evaluate the definite integral at the upper bound x=\pi. Since sin(\pi)=0, the term is 0, but the solution only shows the evaluation at the lower bound (0=0) and omits the subtraction of the lower bound from the upper bound, making the derivation incomplete and logically insufficient to conclude the result is 0.
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.