Question

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

Options

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

• log |sin x + cos x| + C

• log |sin x - cos x| + C

• ${\displaystyle \frac{1}{{(\sin x+\cos x)}^2}}$

Answer 1

${\displaystyle \int \frac{\cos 2x}{{(\sin x+\cos x)}^2}dx}$ is equal to log |sin x + cos x| + C.

Explanation:

Let ${\displaystyle I=\frac{\cos 2x}{\sin x+\cos x}dx}$

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

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

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

put cos x + sin x = t 

⇒ (-sin x + cos x)dx = dt

${\displaystyle =\int \frac{dt}{t}=\log t+C}$

= log |sin x + cos x| + C