Disguised as Integral

It is given that $\displaystyle I = \int \limits_0^{\frac{\pi }{2}} \frac{\text{ d}x}{(1 + \sin x)(1+\sin x - \cos x)}$ .

Rearranging a little, we have,
$\displaystyle I = \int \limits_0^{\frac{\pi }{2}} \frac{\sec ^2 x \text{ d}x}{(\sec x + \tan x)(\sec x+\tan x - 1)}$ .

$\displaystyle \Rightarrow I = \int \limits_0^{\frac{\pi }{2}} \frac{1}{(\sec x + \tan x)^2}. \frac{1}{\bigg(1- \frac{1}{\sec x+\tan x} \bigg)}.\sec ^2 x \text{ d}x$

The expression inside the integral reminds of sum of a geometric progression.Therefore,

$\displaystyle \Rightarrow I = \int \limits_0^{\frac{\pi }{2}} \sum_{n=2}^\infty \frac{1}{(\sec x + \tan x)^n}. \sec ^2 x \text{ d}x$

$\displaystyle \Rightarrow I = \sum_{n=2}^\infty \int \limits_0^{\infty} \frac{\text{ d}y}{(y + \sqrt{1 + y^2})^n}$

Now, I will directly use a result without a proof which is $\displaystyle \int \limits_0^{\infty} \frac{\text{ d}y}{(y + \sqrt{1 + y^2})^n} = \frac{n}{n^2 - 1}$

$\displaystyle \Rightarrow I = \sum_{n=2}^\infty \frac{n}{n^2 - 1}$

By Partial Fractions, $\displaystyle \Rightarrow I = \bigg(\sum_{n=1}^\infty \frac{1}{n} \bigg) - \frac{3}{4}$ .

Using $\displaystyle \lim_{n \to \infty} \bigg( \sum_{r=1}^n \frac{1}{r} \bigg) - ln \text{ n} = 0.5772$ , we get $\displaystyle \text{ln } L =\lim_{n \to \infty} \bigg( \sum_{r=1}^n \frac{1}{r} \bigg) - 0.75 - \text{ln } n = -0.173$ .