Integration Of Rational Trigonometric Functions
Using the Weierstrass half-angle substitution, evaluate the indefinite integral of one divided by the quantity one plus sine x across its valid domain.
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About This Question
- Subject
- mathematics
- Chapter
- integral calculus
- Topic
- integration of rational trigonometric functions
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
−1+tan(x/2)2+C
Rational functions of sine and cosine yield universally to the Weierstrass half-angle substitution t = \tan(x/2), which expresses every trigonometric quantity as an ordinary rational function of t. The standard identities are \sin x = \frac{2t}{1+t^2} and dx = \frac{2,dt}{1+t^2}. Using them, the denominator becomes 1 + \sin x = \frac{1 + t^2 + 2t}{1+t^2} = \frac{(1+t)^2}{1+t^2}, a perfect square over the standard factor. The integral then simplifies to \int \frac{1+t^2}{(1+t)^2}\cdot\frac{2,dt}{1+t^2} = \int \frac{2,dt}{(1+t)^2} = -\frac{2}{1+t} + C = -\frac{2}{1+\tan(x/2)} + C. Option \tan(x/2) drops the entire denominator transformation. Option \ln|1+\sin x| wrongly treats the denominator as if its derivative were the numerator. Option with 1 - \tan(x/2) flips a sign inside the perfect square. The power of this method is that it rationalizes any trigonometric quotient mechanically, turning a problem that resists ordinary substitution into a routine polynomial integration. As a final plausibility check, differentiating -2/(1+\tan(x/2)) through the quotient and chain rules and then simplifying reproduces 1/(1+\sin x) exactly, a result valid wherever 1 + \sin x \neq 0, which excludes only the isolated points where the half-angle tangent blows up.
This hard difficulty mathematics question is from the chapter integral calculus, covering the topic of integration of rational trigonometric functions. It appeared in the 2025 exam.
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