If $I_1 = \int_0^{\pi / 4} \sin^2 x \, dx$ and $I_2 = \int_0^{\pi / 4} \cos^2 x \, dx$,then,

If $\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x$,then the value of $\sqrt{7} \tan \left(\frac{2 \alpha \sqrt{7}}{\pi}\right)$ is $....$. (Here,the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.)

Let $f: [2, 5] \to [2, 5]$ be a bijective function such that $\frac{d}{dx}(f^{-1}(x)) > 0$ for all $x \in [2, 5]$. Then $\int_{2}^{5} (f(x) + f^{-1}(x)) dx$ is

$\int \limits_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x$ is equal to:

$\int_0^{\pi / 2} \frac{\sin ^3 x \cos x \, dx}{\sin ^4 x+\cos ^4 x} = ?$

The absolute value of $\frac{\int_{0}^{\pi/2} (x \cos x + 1) e^{\sin x} dx}{\int_{0}^{\pi/2} (x \sin x + 1) e^{\cos x} dx}$ is equal to -