$\int_0^{2a} f(x) dx - \int_a^{2a} f(x) dx =$

If ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta ,} $ then for any positive integer $n,$ the value of $n({I_{n - 1}} + {I_{n + 1}})$ is

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 $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.)

If $\int\limits_0^1 \frac{\ln x}{\sqrt{1 - x^2}} dx = k \int\limits_0^\pi \ln(1 + \cos x) dx$,then the value of $k$ is:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$ is equal to

$\int_0^\pi x \sin x \, dx = $