Integrals of powers of sine and cosine
Problem 4.864 · medium
Evaluate \( \displaystyle \int_0^{\pi} \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx \).
- The power of cosine is odd: keep one cos x for du, write the rest as (1 − sin²x)^1, and let u = sin x.
- \[ \frac{d}{d u} \left(- \frac{u^{5}}{5} + \frac{u^{3}}{3}\right) = - u^{4} + u^{2} \]∫ -u**4 + u**2 du = -u**5/5 + u**3/3.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{\sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}\right) = \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} \]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. Although the result happens to be 0, the step 'Evaluate from 0' is incomplete and does not demonstrate the application of the Fundamental Theorem of Calculus over the full interval [0, pi].
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to evaluate the antiderivative at the upper limit x = pi. Although the result happens to be 0, the step 'Evaluate from 0' is incomplete and does not demonstrate the application of the Fundamental Theorem of Calculus over the full interval [0, pi].gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies the standard technique for integrating odd powers of cosine. The antiderivative is correct, and the evaluation at the bounds yields 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 2026-10-10 with SymPy 1.14.0.