Question

Evaluate the definite integral ${\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{9+16\sin 2x}dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{9+16\sin 2x}dx$
Also, let $\sin x-\cos x=t\Rightarrow (\cos x+\sin x)dx=dt$
When $x=0,t=-1$ and when $x=\frac{\pi }{4},t=0$
$\Rightarrow (\sin x-\cos x{)}^2=t^2$
$\Rightarrow {\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=t^2$
$\Rightarrow 1-\sin 2x=t^2$
$\Rightarrow \sin 2x=1-t^2$
$\therefore I={\int }_{-1}^0\frac{dt}{9+16(1-t^2)}$
$={\int }_{-1}^0\frac{dt}{9+16-16t^2}$
$={\int }_{-1}^0\frac{dt}{25-16t^2}={\int }_{-1}^0\frac{dt}{(5{)}^2-(4t{)}^2}$
$=\frac{1}{4}{\left[\frac{1}{2\left(5\right)}\log \left|\frac{5+4t}{5-4t}\right|\right]}_{-1}^0$
$=\frac{1}{40}\left[\log \right(1)-\log \left|\frac{1}{9}\right|]$
$=\frac{1}{40}\log 9$