If $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} \mathrm{~d} x$, then $\int_0^{2I} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x$ equals :

The value of $\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x$ is

Let for $f(x)=7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x, \quad \mathrm{I}_1=\int_0^{\pi / 4} f(x) \mathrm{d} x$ and $\mathrm{I}_2=\int_0^{\pi / 4} x f(x) \mathrm{d} x$. Then $7 \mathrm{I}_1+12 \mathrm{I}_2$ is equal to :

The integral $$\int_\limits{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x$$ is equal to