$$\int_{0}^{\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx$$
My Try: Let
$$\tag1 I = \int_{0}^{\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx.$$
Put $x=\frac{\pi}{2}-x$. Using $\int_{0}^{a}f(x)\,\mathrm dx = \int_{0}^{a}f(a-x)\,\mathrm dx$.
So $$\tag2 I = \int_{0}^{\frac{\pi}{2}...
This question was closed yesterday, due to lack of effort. But the comments by OP show that he wasn't successful in his attempts. This needs to be re-opened. Is that possible?
$$\int_{0}^{\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx$$
My Try: Let
$$\tag1 I = \int_{0}^{\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx.$$
Put $x=\frac{\pi}{2}-x$. Using $\int_{0}^{a}f(x)\,\mathrm dx = \int_{0}^{a}f(a-x)\,\mathrm dx$.
So $$\tag2 I = \int_{0}^{\frac{\pi}{2}...