$\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$ is equal to

If for non-zero $x,$ $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5,$ where $a \ne b,$ then $\int_1^2 {f(x)\,dx = } $

The value of $\int_{-1}^1 \sin^5 x \cos^4 x \, dx$ is

$\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{2}{3}} x}{\sin^{\frac{2}{3}} x + \cos^{\frac{2}{3}} x} dx =$

If $I$ is the greatest of the definite integrals
${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then

If $I_{n} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \, dx$,where $n$ is a positive integer,then $I_{10} + I_{8}$ is equal to