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Integrals of powers of sine and cosine

Problem 4.567 · medium

Evaluate \( \displaystyle \int_0^{\frac{\pi}{2}} \sin^{5}{\left(x \right)}\, dx \).

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
\[ \frac{8}{15} \]
Evaluate from 0.✓ Proved

Answer \( \frac{8}{15} \)

Lines: 3 proved, 1 reviewed. 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	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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the standard substitution method for odd powers of sine, and the algebraic steps and final evaluation are verified as correct.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the standard substitution method for odd powers of sine, and the algebraic steps and final evaluation are verified as correct.
gpt-oss:20b: fail (misleading) 2026-10-07 — The solution omits the crucial step of applying the limits after the substitution; a student may be misled into thinking the evaluation is trivial and may not learn how to handle the bounds correctly.
qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the standard substitution method for odd powers of sine. The algebraic steps and final evaluation are correct, and the CAS checks confirm the antiderivative and result.

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.