\( \int\displaystyle\frac{2\cos 2x\,dx}{(\sin x+\cos x)^2} \) is equal to:(where C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2\log|\sin x+\cos x|+\mathrm{C} \)
The numerator \( \cos 2x \) factorises as a difference of squares:
\( \displaystyle \cos 2x=\cos^2x-\sin^2x=(\cos x-\sin x)(\cos x+\sin x) \)
\( \displaystyle \int\frac{2\cos 2x}{(\sin x+\cos x)^2}\,dx=2\int\frac{\cos x-\sin x}{\sin x+\cos x}\,dx \)
The numerator is the derivative of the denominator. Put \( t=\sin x+\cos x \), \( dt=(\cos x-\sin x)\,dx \):
\( \displaystyle 2\int\frac{dt}{t}=2\log|t|+\mathrm{C} \)
\( \displaystyle =2\log|\sin x+\cos x|+\mathrm{C} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{-4}{\sin x+\cos x}+\mathrm{C} \) — Incorrect. Differentiating it gives \( \frac{4(\cos x-\sin x)}{(\sin x+\cos x)^2} \), which is not the integrand \( \frac{2(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^2} \).
Option (2) \( 2\log|\sin x-\cos x|+\mathrm{C} \) — Incorrect. The derivative of \( \sin x-\cos x \) is \( \cos x+\sin x \), not \( \cos x-\sin x \), so the logarithm must contain \( \sin x+\cos x \).
Option (3) \( \log|\sin x+\cos x|+\mathrm{C} \) — Incorrect. The factor 2 in the numerator has been dropped.