If $f\left(x\right)=$$\underset{n\to \infty }{Lt}{e}^{x\tan \left(\frac{1}{n}\right)\log \left(\frac{1}{n}\right)}$ and $\int \frac{f\left(x\right)}{\sqrt[3]{{\sin }^{11}x\cos x}}dx=g\left(x\right)+c$ then 

$f\left(x\right)=\underset{n\to \infty }{Lt}{e}^{x\tan \frac{1}{n}\log \frac{1}{n}}$

$e^{\underset{n\to \infty }{Lt}x\left(\frac{\log 1/n}{\cot 1/n}\right)}$

$$\begin{gathered} =e^{\underset{n\to \infty }{Lt}x\left(\frac{\frac{1}{(1/n)}\times \frac{-1}{n^2}}{\left(-\cos e{c}^2\frac{1}{4}\right)(-1/n^2)}\right) } \\ \left(by L-Hospital rule\right) \end{gathered}$$

$=e^{\underset{n\to \infty }{Lt}\left(x\left(\frac{{\displaystyle -{\sin }^{2}1/n}}{{\displaystyle 1/n}}\right)\right)}$

$=e^{x(-1)\left(0\right)}=e^0=1$

$Now \int \frac{f\left(x\right)}{\sqrt[3]{{\sin }^{11}xcosx}}=\int \frac{1}{{\left(\frac{{\sin }^{11}x}{{\cos }^{11}x}\times {\cos }^{12}x\right)}^{1/3}}dx$

$=\int \frac{{\sec }^{4}x}{{\left(\tan x\right)}^{11/3}}dx$

$=\int \frac{(1+{\tan }^{2}x)}{{\left(\tan x\right)}^{11/3}}.{\sec }^{2}xdx$

$=\int \left(\frac{1+t^2}{t^{11/3}}\right)dt$

$=\int \left(t^{-11/3}+t^{-5/3}\right)dt=\frac{t^{-8/3}}{-8/3}+\frac{t^{-2/3}}{-2/3}+C$

$=\frac{-3}{8t^{8/3}}-\frac{3}{2t^{2/3}}+c$

$\therefore g\left(x\right)=\frac{-3}{8{\tan }^{8/3}x}-\frac{3}{2\left({\cos }^{2/3}x\right)}$

$\therefore g(\pi /4)=\frac{-3}{8\left(1\right)}-\frac{3}{2}=\frac{-3/-12}{8}=\frac{-15}{8}$