Question

Evaluate;-
$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin x+\cos x}{16+9\sin 2x}dx$

Answer 1

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

Noe let $y$ simplify

$16+9\sin 2x$

$25-9+9\sin 2x$

$25-9(1-\sin 2x)$

$25-9({\sin }^{2}x+{\cos }^{2}x-\sin 2x)$

$=25-9(\sin x-\cos x{)}^2$

${\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{25-9(\sin x-\cos x{)}^2}dx$

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

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

${\int }_0^{\frac{\pi }{4}}\frac{dt}{25-9t^2}=\frac{1}{9}{\int }_0^{\frac{\pi }{4}}\frac{dt}{{\left(\frac{5}{3}\right)}^2-t^2}$

$=\frac{1}{9}\times \frac{1}{2\times \frac{5}{3}}{\left[\frac{\log |\frac{5}{3}+t|}{\log |\frac{5}{3}-t|}\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{30}{\left[\frac{\log |\frac{5}{3}+(\sin x-\cos x)|}{\log |\frac{5}{3}-(\sin x-\cos x)|}\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{30}\times \frac{{\log }^{2}2/3}{\log 8/3}$

$=\frac{1}{30}\frac{\log 2-\log 3}{\log 8-\log 3}$