Question

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

Answer 1

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

express deno. in terms of sinx-cosx.

$sin2x=1-(sinx-cosx{)}^2$

$\Rightarrow {\int }_0^{\pi /4}\frac{sinx+cosx}{9+16(1-(sinx-cosx{)}^2)}dx$

$\Rightarrow {\int }_0^{\pi /4}\frac{sinx+cosx}{25-16(sinx-cos{x}^2)}dx$

Put $sinx-cosx=t$

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

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

$=\frac{1}{40}log9.$