Question

$\int \frac{\sin x+\cos x}{9+16\sin 2x}dx$

Answer 1

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

$(\cos x+\sin x)dx=dt$ ${(\sin x-\cos x)}^2=t^2$

Squaring on both sides

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

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

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

$\int \frac{\sin x+\cos x}{9+16\sin 2x}dx$

$\int \frac{dt}{9+16\sin 2x}=\int \frac{dt}{25-16t^2}=\frac{1}{16}\int \frac{dt}{{\left(\frac{5}{4}\right)}^2-16t^2}$

$=\frac{1}{16}\times \frac{1}{2\left(\frac{5}{4}\right)}\log \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|+c$ $=\frac{1}{40}\log \left|\frac{5+4\left(\sin x\cos x\right)}{5-4\left(\sin x\cos x\right)}\right|+c$