Evaluate improper integral $\int_0^{\pi/2}(\sec^2x-\sec x \tan x)dx$

Question

Evaluate the improper integral $$\int_0^{\pi/2}(\sec^2x-\sec x \tan x)dx$$

I got $$\int_0^{\pi/2}(\sec^2x-\sec x \tan x)dx=\int_0^{\pi/2} \frac{1-\sin x}{\cos^2x}dx$$

The singularity occurs at the upper bound $\pi/2$. But, I don't know how to deal with it to evaluate the integral.

I would greatly appreciate any help.

Answer 1

$$\lim_{z\to \pi/2^-}\int_0^z(\sec^2x-\sec x \tan x)\,dx=\lim_{z\to \pi/2^-}[\tan z-\sec z+1]=\lim_{z\to \pi/2^-}\frac{\sin x+\cos x-1}{\cos x}=1$$

Answer 2

Hint: First of all, $~\tan'x=1+\tan^2x=\sec^2x.~$ Secondly, $~\cos'x=-\sin x.$

Answer 3

To avoid dealing with the singularity at the upper bound $x\to \frac\pi2$, continue manipulating the integrand as follows

\begin{align} &\int_0^{\pi/2}(\sec^2x-\sec x \tan x)dx\\=&\int_0^{\pi/2} \frac{1-\sin x}{\cos^2x}dx= \int_0^{\pi/2} \frac{1-\cos x}{\sin^2x}dx\\ =& \ \frac12 \int_0^{\pi/2} \frac{1}{\cos^2\frac x2}dx =\tan\frac x2\bigg|_0^{\pi/2}=1 \end{align}