Integrals of powers of sine and cosine
Problem 4.416 · medium
Evaluate \( \displaystyle \int_0^{\pi} \sin^{5}{\left(x \right)}\, dx \).
- The power of sine is odd: keep one sin x for du, write the rest as (1 − cos²x)^2, and let u = cos x (du = −sin x dx).
- \[ \frac{d}{d u} \left(- \frac{u^{5}}{5} + \frac{2 u^{3}}{3} - u\right) = - u^{4} + 2 u^{2} - 1 \]∫ -u**4 + 2*u**2 - 1 du = -u**5/5 + 2*u**3/3 - u.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}\right) = \sin^{5}{\left(x \right)} \]Differentiating the answer returns the integrand.✓ Proved
- \[ \frac{16}{15} \]Evaluate from 0.✓ Proved
Answer \( \frac{16}{15} \)
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 claims to evaluate the definite integral but only provides the indefinite antiderivative and checks its derivative. It fails to perform the actual evaluation of the antiderivative at the bounds 0 and pi, which is the core of the problem.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims to evaluate the definite integral but only provides the indefinite antiderivative and checks its derivative. It fails to perform the actual evaluation of the antiderivative at the bounds 0 and pi, which is the core of the problem.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the method for integrating an odd power of sine, performs the substitution and integration accurately, and evaluates the definite integral to the correct result.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.