Question

Integrate $\frac{\cos x-\sin x}{1+\sin 2x}$

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

Answer 1

Answer: (A)

$\frac{-1}{\sin x+\cos x}+C$

$I=\int \left(\frac{cosx-sinx}{1+2sinxcosx}\right)dx=\int \left(\frac{cosx-sinx}{si{n}^{2}x+co{s}^{2}x+2sinxcosx}\right)dx=\int \frac{cosx-sinx}{(sinx+cosx{)}^2}dxletsinx+cosx=t(cosx-sinx)dx=dt\therefore =\int \left(\frac{dt}{t^2}\right)=\left(\frac{-1}{t}\right)+c=\left(\frac{-1}{(sinx+cosx)}\right)+c$