Question

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

Answer 1

Let $\sin x-\cos x=t\Rightarrow (\cos x+\sin x)dx=dtx=0,t=-1x=\frac{\pi }{4},t=0\Rightarrow {(\sin x-\cos x)}^2=t^2\Rightarrow 1-2\sin x\cos x=t^2\Rightarrow 1-\sin 2x=t^2\Rightarrow \sin 2x=1-t^2\therefore I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos xdx}{9+16\sin 2x}={\int }_{-1}^0\frac{dt}{9+16(1-t^2)}={\int }_{-1}^0\frac{dt}{9+16-16t^2}={\int }_{-1}^0\frac{dt}{25-16t^2}={\int }_{-1}^0\frac{dt}{5^2-{\left(4t\right)}^2}=\frac{1}{4}{\left[\frac{1}{2\left(5\right)}\log \left|\frac{5+4t}{5-4t}\right|\right]}_1^0=\frac{1}{40}[\log \left(1\right)-\log \left(\frac{1}{9}\right)]=\frac{1}{40}\log \left(9\right)$