By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$.

The value of the integral $\int_{-1}^{1} \frac{d}{dx} \left( \tan^{-1} \frac{1}{x} \right) dx$ is

$e^{\int_0^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x}=$

If $I = \int\limits_0^{\frac{\pi}{2}} \ln(\sin x) dx$,then $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx =$

Let $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$ for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the correct expression$(s)$ is(are):
$(A) \int_0^{\pi/4} x f(x) dx = \frac{1}{12}$
$(B) \int_0^{\pi/4} f(x) dx = 0$
$(C) \int_0^{\pi/4} x f(x) dx = \frac{1}{6}$
$(D) \int_0^{\pi/4} f(x) dx = 1$

Let $I_{n} = \int_{1}^{e} x^{19}(\log |x|)^{n} dx$,where $n \in N$. If $(20) I_{10} = \alpha I_{9} + \beta I_{8}$,for natural numbers $\alpha$ and $\beta$,then $\alpha - \beta$ is equal to ..... .