Integrals of powers of sine and cosine
Problem 4.1016 · medium
Evaluate \( \displaystyle \int_0^{\pi} \sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx \).
- 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
- \[ 0 \]Evaluate from 0.✓ Proved
Answer \( 0 \)
✓ 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 substitution method for odd powers of cosine and sine. The antiderivative is verified, and the evaluation at the bounds (both yielding 0) is correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly applies the substitution method for odd powers of cosine and sine. The antiderivative is verified, and the evaluation at the bounds (both yielding 0) is correct.gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly applies the substitution method for odd powers of cosine and sine, and the evaluation of the definite integral is 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 2026-10-13 with SymPy 1.14.0.