Question

${\int }_0^{\pi /4}\frac{(sinx+cosx)}{9+16sin2x}dx$

Answer 1

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

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

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

Again ${(\sin x-\cos x)}^2=t^2$

$\Rightarrow {\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=t^2$

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

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

When $x=0\Rightarrow t=\sin 0-\cos 0=-1$

When $x=\frac{\pi }{4}\Rightarrow t=\sin \frac{\pi }{4}-\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0$

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

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

$={\int }_{-1}^0\frac{dt}{25-16t^2}$

$=\frac{1}{16}{\int }_{-1}^0\frac{dt}{\frac{25}{16}-t^2}$

$=\frac{1}{16}{\left[\frac{1}{2\times \frac{5}{4}}\log \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|\right]}_{-1}^0$

$=\frac{1}{4}{\left[\frac{1}{10}\log \left|\frac{5+4t}{5-4t}\right|\right]}_{-1}^0$

$=\frac{1}{40}[\log \left|\frac{5+0}{5-0}\right|-\log \left|\frac{5-4}{5+4}\right|]$

$=\frac{1}{40}[\log 1-\log \left(\frac{1}{9}\right)]$

$=\frac{1}{40}[0+\log 9]$

$=\frac{\log 9}{40}$