Question

$\int \frac{1}{1+\cos 3x}dx$

Answer 1

$$\begin{gathered} \int \frac{dx}{1+\cos 3x} \\ =\int \frac{\left(1-\cos 3x\right)}{\left(1+\cos 3x\right)\left(1-\cos 3x\right)}dx \\ =\int \left(\frac{1-\cos 3x}{1-{\cos }^{2}3x}\right)dx \\ =\int \left(\frac{1-\cos 3x}{{\sin }^{2}3x}\right)dx \\ =\int {\mathrm{cosec}}^{2}3xdx-\int \mathrm{cosec}3x\cot 3xdx \\ =-\frac{\cot 3x}{3}+\frac{\mathrm{cosec}3x}{3}+\mathrm{C} \\ =\frac{1}{3}\left[\mathrm{cosec}3x-\cot 3x\right]+\mathrm{C} \\ =\frac{1}{3}\left[\frac{1}{\sin 3x}-\frac{\cos 3x}{\sin 3x}\right]+\mathrm{C} \\ =\frac{1}{3}\left[\frac{1-\cos 3x}{\sin 3x}\right]+\mathrm{C} \end{gathered}$$