Question

$\int \frac{\cos 2x}{(\sin x+\cos x{)}^2}dx$ is equal to

• A. $\frac{-1}{(\sin x+\cos x)}+C$
• B. $\log |\sin x+\cos x|+C$
• C. $\log |\sin x-\cos x|+C$
• D. $\frac{1}{(\sin x+\cos x{)}^2}$

Answer 1

Answer: (B)

$\log |\sin x+\cos x|+C$

Let $I=\int \frac{\cos 2x}{(\sin x+\cos x{)}^2}dx$

$I=\int \frac{{\cos }^{2}x-{\sin }^{2}x}{(\cos x+\sin x{)}^2}dx$

$=\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x{)}^2}dx$

$=\int \frac{\cos x-\sin x}{\cos x+\sin x}dx$

Let $\cos x+\sin x=t\Rightarrow (\cos x-\sin x)dx=dt$

$\therefore I=\int \frac{dt}{t}=\log |t|+C$

$=\log |\cos x+\sin x|+C$

Hence, the correct Answer is B.