Question

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

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

Answer 1

Answer: (B)

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

${\int }_0^{\pi /2}\frac{sinx+cosx}{3+sin2x}dx.$

${\int }_0^{\pi /2}\frac{sinx+cosx}{4-(sinx-cosx{)}^2}$

$dx={\int }_0^{\pi /2}\frac{d(sinx-cosx)}{(2{)}^2-(sinx-cosx{)}^2}$

$=\frac{1}{4}\log {\left|\frac{2+(sinx-cosx)}{2-(sinx-cosx)}\right|}_0^{\pi }$

$=\frac{1}{4}(\log \left|\frac{2+0}{2-0}\right|-\log \left|\frac{2-1}{2+1}\right|)$

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

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