Integrals of powers of sine and cosine
Problem 4.571 · medium
Evaluate \( \displaystyle \int_0^{\frac{\pi}{2}} \sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx \).
- The power of cosine is odd: keep one cos x for du, write the rest as (1 − sin²x)^1, and let u = sin x.Reviewed
- \[ \frac{d}{d u} \left(- \frac{u^{6}}{6} + \frac{u^{4}}{4}\right) = - u^{5} + u^{3} \]∫ -u**5 + u**3 du = -u**6/6 + u**4/4.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{\sin^{6}{\left(x \right)}}{6} + \frac{\sin^{4}{\left(x \right)}}{4}\right) = \sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)} \]Differentiating the answer returns the integrand.✓ Proved
- \[ 1 \cdot \frac{1}{12} = \frac{1}{12} \]Evaluate from 0.✓ Proved
Answer \( \frac{1}{12} \)
✓ 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 applies the standard technique for integrating odd powers of sine and cosine. The algebraic steps and final evaluation are correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the standard technique for integrating odd powers of sine and cosine. The algebraic steps and final evaluation are correct.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the standard technique for integrating odd powers of sine and 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.