The value of the integral $\int_{\pi / 6}^{\pi / 2} \left( \frac{1+\sin 2x+\cos 2x}{\sin x+\cos x} \right) dx$ is equal to

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $

The solution for $x$ of the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$ is

If $\left( \int_{0}^{a} x \, dx \right) \le (a + 4)$,then

Let $(2^{1-a} + 2^{1+a})$, $f(a)$, $(3^a + 3^{-a})$ be in $A$.$P$. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$ is:

$\int_1^2 \frac{1}{x^2} e^{-\frac{1}{x}} \, dx = $