Question

The value of the integral ${\int }_0^{\pi /4}\frac{\sin x+\cos x}{3+\sin 2x}dx$ 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

The correct option is D $\frac{1}{4}{\log }_{e}3$

Let

$I={\int }_0^{\pi /4}\frac{\sin x+\cos x}{3+\sin 2x}dx$

$={\int }_0^{\pi /4}\frac{\sin x+\cos x}{3+2\sin x\cos x}dx$

$={\int }_0^{\pi /4}-\frac{\sin x+\cos x}{(\sin x-\cos x{)}^2-4}dx$ ...... (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,

$I=-{\int }_{-1}^0\frac{dt}{t^2-4}=-\frac{1}{4}{\left[\log \left|\frac{t-2}{t+2}\right|\right]}_{-1}^0$

$=-\frac{1}{4}(\log 1-\log 3)$

$=-\frac{1}{4}(0-\log 3)=\frac{1}{4}\log 3$