Question

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

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

Answer 1

The correct option is B $\ell n(\sin x+\cos x)+C$

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

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

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

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

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

$(\cos x-\sin x)dx=dt$

$\int \frac{dt}{t}$

$\ln t+C$

$\ln (\sin x+\cos x)+C$

$\therefore \int \frac{\cos 2xdx}{(\sin x+\cos x{)}^2}=\ln (\sin x+\cos x)+C$