Home›Calculus 1›u-substitution›Problem 4.1317

Integral of \( \displaystyle \frac{\sin{\left(2 x - 1 \right)}}{\cos^{2}{\left(2 x - 1 \right)}} \)

Problem 4.1317 · medium Video

Find \( \displaystyle \int \frac{\sin{\left(2 x - 1 \right)}}{\cos^{2}{\left(2 x - 1 \right)}} \, dx \). (Omit the constant of integration.)

\[ \int \frac{\sin{\left(2 x - 1 \right)}}{\cos^{2}{\left(2 x - 1 \right)}}\, dx \]
integralStart with the integral of the given function.✓ Proved
\[ = \int \tan{\left(2 x - 1 \right)} \sec{\left(2 x - 1 \right)}\, dx \]
rewrite simplifyUse the identity sin(u)/cos(u) = tan(u) and 1/cos(u) = sec(u). The expression is already in its simplest form for integration.✓ Proved
\[ = \frac{\sec{\left(2 x - 1 \right)}}{2} \]
antiderivativeThe antiderivative of sec(u)tan(u) is sec(u), and we account for the chain rule with the 1/2 factor.✓ Proved

Answer \( \frac{1}{2 \cos{\left(2 x - 1 \right)}} + C \)

Top: the graph of F = 1/(2cos(2x - 1)). A tangent line slides along it, and its slope is plotted below as it goes, tracing f = sin(2x - 1)/cos(2x - 1)².

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	line 1 is the problem as stated
2	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where cos(2*x - 1) = 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2
3	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2
4	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2
answer	✓ Proved	sympy 1.14.0	final line against the stated answer: simplify(a - b) reduced to 0 undefined where cos(2*x - 1) = 0
answer, a second way	✓ Proved	sympy 1.14.0	SymPy differentiated the stated antiderivative back to the integrand

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the integrand as a derivative of secant via substitution, applying the chain rule factor correctly. The steps are logically sound and the labels are appropriate.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-23 — The solution correctly identifies the integrand as a derivative of secant via substitution, applying the chain rule factor correctly. The steps are logically sound and the labels are appropriate.
gpt-oss:20b: pass 2026-10-23
qwen3.6:27b-mlx: pass 2026-10-22 — The solution correctly identifies the integrand as a derivative of secant via substitution, and the steps adhere to the one-rule-per-step constraint. The final answer matches the stated answer.
gpt-oss:20b: pass 2026-10-22
qwen3.6:27b-mlx: pass 2026-10-22
gpt-oss:20b: pass 2026-10-22

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 gemma4:26b, checked 2026-10-23 with SymPy 1.14.0.