If $\int\left(\frac{1-5 \cos ^2 x}{\sin ^5 x \cos ^2 x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to

Let $f(t)=\int\left(\frac{1-\sin \left(\log _e t\right)}{1-\cos \left(\log _e t\right)}\right) d t, t>1$.

If $f\left(e^{\pi / 2}\right)=-e^{\pi / 2}$ and $f\left(e^{\pi / 4}\right)=\alpha e^{\pi / 4}$, then $\alpha$ equals

Let $\mathrm{I}(x)=\int \frac{3 d x}{(4 x+6)\left(\sqrt{4 x^2+8 x+3}\right)}$ and $\mathrm{I}(0)=\frac{\sqrt{3}}{4}+20$. If

$\mathrm{I}\left(\frac{1}{2}\right)=\frac{a \sqrt{2}}{b}+\mathrm{c}$, where $a, b, \mathrm{c} \in \mathrm{N}, \operatorname{gcd}(a, b)=1$, then $a+b+c$ is equal to :

Let $f(x)=\int \frac{\left(2-x^2\right) \cdot \mathrm{e}^x}{(\sqrt{1+x})(1-x)^{3 / 2}} \mathrm{~d} x$. If $f(0)=0$, then $f\left(\frac{1}{2}\right)$ is equal to: