Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\left(\sin x+\cos x\right)dx}{9+16\sin 2x} \\ \mathrm{Let}\sin x-\cos x=t.....\left(1\right) \\ \mathrm{Diff}\mathrm{both}\mathrm{sides} \\ \left(\cos x+\sin x\right)dx=dt \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right),\mathrm{we}\mathrm{get} \\ {\left(\sin x-\cos x\right)}^2=t^2 \\ \Rightarrow {\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=t^2 \\ \Rightarrow 1-t^2=2\sin x\cos x \\ \Rightarrow 1-t^2=\sin \left(2x\right) \\ \therefore I=\int \frac{dt}{9+16\left(1-t^2\right)} \\ =\int \frac{dt}{25-16t^2} \\ =\frac{1}{16}\int \frac{dt}{{\displaystyle \frac{25}{16}-t^2}} \\ =\frac{1}{16}\int \frac{dt}{{\left({\displaystyle \frac{5}{4}}\right)}^2-t^2} \\ =\frac{1}{16}\times \frac{1}{2\times {\displaystyle \frac{5}{4}}}\log \left|\frac{{\displaystyle \frac{5}{4}+t}}{{\displaystyle \frac{5}{4}-t}}\right|+C \\ =\frac{1}{40}\log \left|\frac{5+4\left(\sin x-\cos x\right)}{5-4\left(\sin x-\cos x\right)}\right|+C \end{gathered}$$