Question

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

• A. $\log 2$
• B. $\log 3$
• C. $\frac{1}{4}\log 3$
• D. $\frac{1}{8}\log 3$

Answer 1

The correct option is C $\frac{1}{4}\log 3$

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

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

substitute $u=\sin x-\cos x\Rightarrow (\sin x+\cos x)dx=dt$

$x(0,\frac{\pi }{4})\to t(-1,0)$

$={\int }_{-1}^0\frac{dt}{4-t^2}$

$={\int }_{-1}^0\frac{dt}{2^2-t^2}$

$=\frac{1}{2\times 2}{\left[\ln \frac{2+t}{2-t}\right]}_{-1}^0$

$=\frac{1}{4}\{\ln 1-\ln \frac{1}{3}\}$

$=\frac{1}{4}\ln 3$