Question

\(\int \frac{cos x - sin x}{1 + sin2x} dx\) is equal to:

• A. \(\frac{1}{sinx + cosx}c\), where C is a constant.
• B. \(\frac{- 1}{sinx + cosx}c\), where C is a constant.
• C. \(\frac{1}{cos x - sinx}c\), where C is a constant.
• D. \(\frac{2}{sinx + cosx}c\), where C is a constant.

Answer 1

The Correct Option is B

To solve the integral \(\int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx\), we start by simplifying the expression in the denominator. Recall that \(\sin 2x = 2 \sin x \cos x\), so the denominator can be rewritten as: \[1 + \sin 2x = 1 + 2\sin x \cos x = (\sin x + \cos x)^2\] Substituting, the integral becomes: \[\int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx\] Let's set \(u = \sin x + \cos x\), then \(du = (\cos x - \sin x)dx\). The differential \(du\) matches the numerator, thereby simplifying the integral to: \[\int \frac{1}{u^2} \, du\] The antiderivative of \(\frac{1}{u^2}\) is \(-\frac{1}{u}\), thus integrating gives: \[-\frac{1}{u} + C\] Back substituting \(u = \sin x + \cos x\), the solution to the integral is: \[-\frac{1}{\sin x + \cos x} + C\] Hence, the answer is \(\frac{-1}{\sin x + \cos x} + C\), which matches the given option:

\(\frac{- 1}{\sin x + \cos x}c\), where C is a constant.