Question

${\int }_0^{\pi /2}\frac{\cos 2x}{{(\sin x+\cos x)}^2}dx=......$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $0$
• D. $-\frac{\pi }{4}$

Answer 1

Answer: (C)

$0$

$I={\int }_0^{\pi /2}\frac{\cos 2x}{{(\sin x+\cos x)}^2}dx$
$={\int }_0^{\pi /2}\frac{\cos 2x}{1+2\sin x\cos x}dx(\therefore co{s}^{2}x+si{n}^{2}x=1)$

$={\int }_0^{\pi /2}\frac{\cos 2x}{1+\sin 2x}dx$
$\frac{1}{2}{\int }_0^{\pi /2}\frac{\frac{d}{dx}(1+\sin 2x)}{1+\sin 2x}dx(\therefore substitutecos2x=1+sin2x)$

$=\frac{1}{2}{\left[\log |1+\sin 2x|\right]}_0^{\pi /2}=\frac{1}{2}[\log 1-\log 1]$
$\therefore I=0$