Question

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

Answer 1

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

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

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

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

When $x=0,t=-1$ and when $x=\frac{\pi }{4},t=0$

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

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

= $\frac{1}{9}{\int }_{-1}^0\frac{dt}{{\left(\frac{5}{3}\right)}^2-t^2}=\frac{1}{9}\times \frac{1}{2\times \frac{5}{3}}\times {\left[\log \left|\begin{array}{c}\frac{\frac{5}{3}+t}{\frac{5}{3}-t}\end{array}\right|\right]}_{-1}^0$

= $\frac{1}{30}{\left[\log \left|\begin{array}{c}\frac{5+3t}{5-3t}\end{array}\right|\right]}_{-1}^0=\frac{1}{30}[\log 1-\log \frac{1}{4}]$

= $\frac{1}{30}[\log 1-\log 1+\log 4]$

= $\frac{1}{30}\log 4=\frac{1}{15}\log 2$