Question

$\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\sec xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{4}}\sec x\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{4}}\sec x\frac{\sec x+\tan x}{\sec x+\tan x}\mathit{d}x \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{{\sec }^{2}x+\sec x\tan x}{\sec x+\tan x}\mathit{d}x \\ \mathrm{Put}u=\sec x+\tan x \\ \Rightarrow du={\sec }^{2}x+\sec x\tan xdx \\ \therefore {\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{{\sec }^{2}x+\sec x\tan x}{\sec x+\tan x}dx=\int \frac{du}{u} \\ \Rightarrow I=\left[\log u\right] \\ \Rightarrow I={\left[\log \left(\sec x+\tan x\right)\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ \Rightarrow I=\log \left(\sec \frac{\mathrm{\pi }}{4}+\tan \frac{\mathrm{\pi }}{4}\right)-\log \left(\sec 0+\tan 0\right) \\ \Rightarrow I=\log (\sqrt{2}+1)-\log 1 \\ \Rightarrow I=\log (\sqrt{2}+1) \end{gathered}$$