Question

${\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{16+9\sin 2x}dx$.

Answer 1

${\int }_0^{\pi /4}\frac{\sin x+\cos x}{16+9\sin 2x}dx$

Let $\sin x-\cos x=t$

$\therefore (\cos x+\sin x)dx=dt$

Also, ${\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=dt$

$\Rightarrow 1-\sin 2x=t^2$

$\Rightarrow \sin 2x=1-t^2$

At $x=0,$ $t=-1$

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

$\therefore I={\int }_{-1}^0\frac{dt}{16+9(1-t^2)}$

$={\int }_{-1}^0\frac{dt}{25-9t^2}$

$={\int }_{-1}^0\frac{dt}{5^2-{\left(3t\right)}^2}$

$=\frac{1}{3}{\left[\frac{1}{2\left(3\right)}\log \left|\frac{5+3t}{5-3t}\right|\right]}_{-1}^0$

$=\frac{1}{30}[\log \frac{5}{5}-\log \left|\frac{2}{1}\right|]$

$=\frac{1}{30}\log 4$

Hence, the answer is $\frac{1}{30}\log 4.$