Question

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

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

Answer 1

Answer: (D)

$\frac{1}{4}\log 3$

$I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{3+\sin 2x}dx$
Put $\sin x-\cos x=t$
$\sin x+\cos x=dt$
$1-2\sin x\cos x=t^2$
$2\sin x\cos x=\sin 2x=1-t^2$
$3+\sin 2x=4-t^2$
$I={\int }_{-1}^0\frac{dt}{4-t^2}$
$I={\left[\frac{1}{4}\log \frac{(2+t)}{(2-t)}\right]}_{-1}^0$
$I=\frac{1}{4}(0-\log \frac{1}{3})=\frac{1}{4}\log 3$