Question

$\int \frac{1}{1+\sec ax}dx$

Answer 1

$$\begin{gathered} \int \frac{dx}{1+\sec ax} \\ =\int \frac{dx}{1+{\displaystyle \frac{1}{\cos ax}}} \\ =\int \frac{\cos ax}{1+\cos ax}dx \\ =\int \left(\frac{1+\cos ax-1}{1+\cos ax}\right)dx \\ =\int \left[\frac{1+\cos ax}{1+\cos ax}-\frac{1}{1+\cos ax}\right]dx \\ =\int dx-\int \frac{dx}{1+\cos ax} \\ =\int dx-\int \frac{1-\cos ax}{\left(1+\cos ax\right)\left(1-\cos ax\right)}dx \\ =\int dx-\int \frac{1-\cos ax}{1-{\cos }^{2}ax}dx \\ =\int dx-\int \frac{1-\cos ax}{{\sin }^{2}ax}dx \\ =\int dx-\int \left(\frac{1}{{\sin }^{2}ax}-\frac{\cos ax}{{\sin }^{2}ax}\right)dx \\ =\int dx-\int \frac{1}{{\sin }^{2}ax}dx+\int \frac{\cos ax}{{\sin }^{2}ax}dx \\ =\int dx-\int \cos e{c}^{2}axdx+\int \cot ax\mathrm{cosec}axdx \\ =x+\frac{\cot ax}{a}-\frac{\mathrm{cosec}ax}{a}+C \\ =x+\frac{1}{a}\left(\cot ax-\mathrm{cosec}ax\right)+C \\ =\int dx-\int \frac{dx}{1+\cos ax} \\ =\int dx-\int \frac{dx}{2{\cos }^2\left({\displaystyle \frac{ax}{2}}\right)}\left[\therefore 1+\cos x=2{\cos }^2\frac{x}{2}\right] \\ =\int dx-\frac{1}{2}\int {\sec }^2\left(\frac{ax}{2}\right)dx \\ =x-\frac{1}{2}\left[\frac{\tan \left({\displaystyle \frac{ax}{2}}\right)}{{\displaystyle \frac{a}{2}}}\right] \\ =x-\frac{\tan \left({\displaystyle \frac{ax}{2}}\right)}{a}+\mathrm{C} \end{gathered}$$