Mathematics

Mathematics

Problem

∫dxcos⁡x−sin⁡x\int \frac{dx}{\cos x - \sin x} is equal to (a) 12log⁡∣tan⁡(x2+3π8)∣+c\frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + c (b) 12log⁡∣cot⁡(x2)∣+c\frac{1}{\sqrt{2}} \log \left| \cot \left( \frac{x}{2} \right) \right| + c (c) 12log⁡∣tan⁡(x2−3π8)∣+c\frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} - \frac{3\pi}{8} \right) \right| + c (d) 12log⁡∣tan⁡(x2−π8)∣+c\frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} - \frac{\pi}{8} \right) \right| + c (a) 12log⁡∣tan⁡(x2+3π8)∣+c\frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + c (b) 12log⁡∣cot⁡(x2)∣+c\frac{1}{\sqrt{2}} \log \left| \cot \left( \frac{x}{2} \right) \right| + c (c) 12log⁡∣tan⁡(x2−3π8)∣+c\frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} - \frac{3\pi}{8} \right) \right| + c (d) 12log⁡∣tan⁡(x2−π8)∣+c\frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} - \frac{\pi}{8} \right) \right| + c
Intuitive explanation

The key is to rewrite the denominator cos⁡x−sin⁡x\cos x - \sin x as a single sinusoidal function using the amplitude-phase form. Factor out 2\sqrt{2} to get 2cos⁡(x+π/4)\sqrt{2}\cos(x + \pi/4), which transforms the integral into a standard form solvable by the Weierstrass substitution t=tan⁡(x/2)t = \tan(x/2). This substitution converts the trigonometric integral into a rational function that yields a logarithm with a specific argument.

Intuitive explanation

The key is to rewrite the denominator cos⁡x−sin⁡x\cos x - \sin x as a single sinusoidal function using the amplitude-phase form. Factor out 2\sqrt{2} to get 2cos⁡(x+π/4)\sqrt{2}\cos(x + \pi/4), which transforms the integral into a standard form solvable by the Weierstrass substitution t=tan⁡(x/2)t = \tan(x/2). This substitution converts the trigonometric integral into a rational function that yields a logarithm with a specific argument.

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