A calculus problem by Jose Sacramento

The question asks one to either use the wonderful result stated prior to it or not!! Let us say that some of us decide not to !! Then, we might say,

Substitute $\large x^{2}$ as $\large tan\theta$

Then,we have $\displaystyle \large \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (tan\theta)^{\frac{1}{2}}d\theta$

which is $\displaystyle \large \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (cot\theta)^{\frac{1}{2}}d\theta$

Let us call this integral $\large I$

$\displaystyle \large 2I = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{cos\theta+sin\theta}{\sqrt{1-(cos\theta-sin\theta)^{2}}}d\theta$

one might take $\displaystyle \large cos\theta-sin\theta$ as some $\alpha$

we have $\large -d\alpha$ as our numerator

simple integration yields $\displaystyle \large \frac{\pi}{2\sqrt{2}}$

Simply using the transformation $x\to x^{1/4}$ , $I=\int_0^\infty \frac{1}{4}\frac{x^{1/4-1}dx}{1+x}=\frac{\pi}{4\sin(\pi/4)}$ which gives the answer $\frac{\sqrt{2}\pi}{4}\approx\boxed{1.11}$ .