Question

Evaluate: $\int \frac{\cos x-\sin x}{1+\sin 2x}dx$

• A. $\frac{1}{\sin x+\cos x}+C$
• B. $-\frac{1}{\sin x+\cos x}+C$
• C. $\frac{2}{\sin 2x+\cos x}+C$
• D. $-\frac{2}{\sin 2x+\cos x}+C$

Answer 1

The correct option is B $-\frac{1}{\sin x+\cos x}+C$

$\int \frac{\sin x-\cos x}{1+\sin 2x}dx=\int \frac{\sin x-\cos x}{{\sin }^{2}x+{\cos }^{2}x+2sinx\cos x}dx=\int \frac{\sin x-\cos x}{{(\sin x+\cos x)}^2}dx$

Put $t=\sin x+\cos x$

$I=\int \frac{1}{t^2}dt=-\frac{1}{t}+C=-\frac{1}{\sin x+\cos x}+C$