Integrals of powers of sine and cosine
Problem 4.564 · hard
Evaluate \( \displaystyle \int \sin^{5}{\left(x \right)}\, dx \).
- The power of sine is odd: keep one sin x for du, write the rest as (1 − cos²x)^2, and let u = cos x (du = −sin x dx).Reviewed
- \[ \frac{d}{d u} \left(- \frac{u^{5}}{5} + \frac{2 u^{3}}{3} - u\right) = - u^{4} + 2 u^{2} - 1 \]∫ -u**4 + 2*u**2 - 1 du = -u**5/5 + 2*u**3/3 - u.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}\right) = \sin^{5}{\left(x \right)} \]Differentiating the answer returns the integrand.✓ Proved
Answer \( - \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)} + C \)
✓ 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 standard technique for integrating odd powers of sine. The algebraic steps and final verification are correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies the standard technique for integrating odd powers of sine. The algebraic steps and final verification are correct.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies the standard technique for integrating odd powers of sine. The algebraic manipulation and final result are verified by differentiation in the subsequent lines.
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.