I was trying to solve the integral $\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx$ and I noticed I can do the following:
$$\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx=\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}} \sec^2(x) -\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}} \tan^2(x) dx $$ $$=\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}} \sec^2(x) -\int_0 ^{\frac{\pi}{4}} \tan^{2+\frac{1}{2}}(x) \sec^2(x)dx + \int_0 ^{\frac{\pi}{4}} \tan^{4+\frac{1}{2}}(x) $$ continue with that and we will get $$\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx=\sum_{k=0}^n \left(\int_0 ^{\frac{\pi}{4}} \tan^{2k+\frac{1}{2}}(x) \sec^2(x)dx \right) + (-1)^{n+1}\int_0 ^{\frac{\pi}{4}} \tan^{2n+2+\frac{1}{2}}(x) dx$$ since for any converging positive sequence $a_n$ $(\sum a_k)^n >\sum (a_k ^n)$ $\int_0 ^{\frac{\pi}{4}} \tan^{n}(x) dx <\left( \int_0 ^{\frac{\pi}{4}} \tan(x) dx \right) ^n \to 0$ so we get $$ \int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx=2 \ \lim_{n \to \infty } \sum_{k=0} ^ n \frac{(-1)^k}{4k+3}=\frac{\pi +\ln{(3-2 \sqrt{2})}}{2 \sqrt{2}} $$
I don't know if my approach is correct or not but even if it is correct I am interested in finding out if there are any other more obvious ways to evaluate $ \sum\limits_{k=0} ^{\infty} \frac{(-1)^k}{4k+3}$ as my approach is very difficult to notice if try to solve $ \sum\limits_{k=0} ^{\infty} \frac{(-1)^k}{4k+3}$ without any mention of the integral.
another question can we generalise this result for all $m \in \mathbb{R}$ st $m>1$ $$ \int_0 ^{\frac{\pi}{4}} \left(\tan{x} \right)^{\frac{1}{m}} dx=m \ \sum_{k=0} ^ {\infty } \frac{(-1)^k}{2mk+m+1} $$