Calculate
$$\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{\frac{{\pi}}{2}}}_{\cssId{lower-bound-mathjax}{0}} \dfrac{1}{1+\tan^n\left(x\right)}\,\cssId{int-var-mathjax}{\mathrm{d}x}$$
where $n$ stands for all natural numbers.
Calculate
$$\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{\frac{{\pi}}{2}}}_{\cssId{lower-bound-mathjax}{0}} \dfrac{1}{1+\tan^n\left(x\right)}\,\cssId{int-var-mathjax}{\mathrm{d}x}$$
where $n$ stands for all natural numbers.
Use the property that $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx$$ Let $$I=\int_{0}^{\pi/2} \frac{\cos^{n}(x)}{\sin^{n}(x)+\cos^{n}(x)} \ dx$$ Then by our property we have $$I=\int_{0}^{\pi/2} \frac{\sin^{n}(x)}{\sin^{n}(x)+\cos^{n}(x)}\ dx$$ Add both