Question

The value of the integral $\underset{0}{\overset{\pi /4}{\int }}\frac{\sin x+\cos x}{3+\sin 2x}dx$ is equal to

• A. ${\log }_{e}2$
• B. ${\log }_{e}3$
• C. $\frac{1}{4}{\log }_{e}2$
• D. $\frac{1}{4}{\log }_{e}3$

Answer 1

Answer: (D)

Let $I=\underset{0}{\overset{\pi /4}{\int }}\frac{\sin x+\cos x}{3+\sin 2x}dx$
$=\underset{0}{\overset{\pi /4}{\int }}\frac{\sin x+\cos x}{3+2\sin x\cos x}dx$
$=\underset{0}{\overset{\pi /4}{\int }}-\frac{\sin x+\cos x}{(\sin x-\cos x{)}^2-4}dx$ ...(i)
Put $\sin x-\cos x=t$
$\Rightarrow (\cos x+\sin x)dx=dt$
when $x=0\Rightarrow t=-1$
and $x=\frac{\pi }{4}\Rightarrow t=0$
$\therefore$ Eq.(i) becomes,
$I=-\underset{-1}{\overset{0}{\int }}\frac{dt}{t^2-4}$
$=-\frac{1}{4}{\left[\log \left|\frac{t-2}{t+2}\right|\right]}_{-1}^0$
$=-\frac{1}{4}(\log 1-\log 3)$
$=-\frac{1}{4}(0-\log 3)=\frac{1}{4}\log 3$