Question

$\int \frac{\cos x}{2+3\sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\cos x}{2+3\sin x}dx \\ \mathrm{Putting}\sin x=t \\ \Rightarrow \cos x=\frac{dt}{dx} \\ \Rightarrow \cos xdx=dt \\ \therefore I=\int \frac{dt}{2+3t} \\ =\frac{1}{3}\ln \left|2+3t\right|+\mathrm{C}\left[\because \int \frac{1}{ax+b}dx=\frac{1}{a}\ln \left|ax+b\right|+\mathrm{C}\right] \\ =\frac{1}{3}\ln \left|2+3\sin x\right|+\mathrm{C}\left[\because \mathit{}t=\sin x\right] \end{gathered}$$