Question

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

• A. $\sin 2x+C$
• B. $-\sin 2x+C$
• C. $\cos 2x+C$
• D. $-\cos 2x+C$

Answer 1

Answer: (A)

$\sin 2x+C$

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

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

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

$=2\int (\cos x-\sin x)(\cos x+\sin x)dx$

$=2\int \cos 2xdx$ $\because (\cos x-\sin x)(\cos x+\sin x)={\cos }^{2}x-{\sin }^{2}x=\cos 2x$

$=2\frac{\sin 2x}{2}$

$=\sin 2x+C$

Hence, the answer is $\sin 2x+C.$