$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x=$

$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

Let $[t]$ denote the greatest integer $\leq t$. Then $\frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$ is equal to

Assertion $(A): \int_{-a}^a f(x) dx = \int_0^a (f(x) + f(-x)) dx$
Reason $(R): \int_a^b f(x) dx = \int_{g(a)}^{g(b)} f(g(u)) g'(u) du$
The correct option among the following is:

$\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x=$

If $2 \int_0^1 \tan^{-1} x \, dx = \int_0^1 \cot^{-1} (1 - x + x^2) \, dx,$ then $\int_0^1 \tan^{-1} (1 - x + x^2) \, dx$ is equal to