Integrals of powers of sine and cosine
Problem 4.566 · medium
Evaluate \( \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}\, dx \).
- The power of cosine is odd: keep one cos x for du, write the rest as (1 − sin²x)^2, and let u = sin x.Reviewed
- \[ \frac{d}{d u} \left(\frac{u^{7}}{7} - \frac{2 u^{5}}{5} + \frac{u^{3}}{3}\right) = u^{6} - 2 u^{4} + u^{2} \]∫ u**6 - 2*u**4 + u**2 du = u**7/7 - 2*u**5/5 + u**3/3.✓ Proved
- \[ \frac{d}{d x} \left(\frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}\right) = \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} \]Differentiating the answer returns the integrand.✓ Proved
- \[ \frac{8}{105} \]Evaluate from 0.✓ Proved
Answer \( \frac{8}{105} \)
✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly identifies the substitution method for an odd power of cosine, performs the integration and back-substitution accurately, and evaluates the definite integral correctly.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies the substitution method for an odd power of cosine, performs the integration and back-substitution accurately, and evaluates the definite integral correctly.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies the standard substitution method for integrals with an odd power of cosine. The algebraic steps and final evaluation are correct.
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.