Question

Evaluate the following definite integral:

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

Answer 1

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

$={\int }_0^{\pi /4}\frac{\sin x+\cos x}{4-1+\sin 2x}dx$

$={\int }_0^{\pi /4}\frac{\sin x+\cos x}{4-(1-\sin 2x)}dx$

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

Let $\sin x-\cos x=t(\cos x+\sin x)dx=dt$

Also,

$t=-1$ at $x=0$

$t=0$ at $x=\frac{\pi }{4}$

Therefore, integral will become

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

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

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

$=\frac{1}{4}(\ln 1+\ln 3)$

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