Integrals of powers of sine and cosine
Problem 4.858 · medium
Evaluate \( \displaystyle \int \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx \).
- The power of cosine is odd: keep one cos x for du, write the rest as (1 − sin²x)^1, and let u = sin x.Reviewed
- \[ \frac{d}{d u} \left(- \frac{u^{5}}{5} + \frac{u^{3}}{3}\right) = - u^{4} + u^{2} \]∫ -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
Answer \( - \frac{\sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3} + C \)
✓ Nihil obstat Lines: 2 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | difference quotients of the answer match the integrand |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the strategy for integrating odd powers of cosine and verifies the result via differentiation. The steps are logically sound and mathematically correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the strategy for integrating odd powers of cosine and verifies the result via differentiation. The steps are logically sound and mathematically correct.gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the standard substitution method for integrals with an odd power of cosine. The algebraic steps and final result 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 2026-10-10 with SymPy 1.14.0.