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  $\int \frac{f\left(x\right)}{\sqrt[3]{{\sin }^{11}x\cdot \cos x}}dx=g\left(x\right)+c$ then 

$f\left(x\right)=\underset{n\to \infty }{Lt}{e}^{x\tan \left(\frac{1}{n}\right)\log \left(\frac{1}{n}\right)}=e^0=1$

$\to \int \frac{f\left(x\right)}{\sqrt[3]{{\sin }^{11}x\cos x}}dx=\int \frac{1}{\sqrt[3]{{\tan }^{11}x{\cos }^{12}x}}dx=\int \frac{1}{{\cos }^{4}x\left({\tan }^{\frac{11}{3}}x\right)}dx$

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

$\tan x=t$

Diff

${\sec }^{2}xdx=dt$

$=\int \left(\frac{1+t^2}{t^{\frac{11}{3}}}\right)dt=\int \left(t^{\frac{-11}{3}}+t^{\frac{-5}{3}}\right)dt$

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

$g\left(x\right)=\frac{-3}{8}{\tan }^{\frac{-8}{3}}x-\frac{3}{2}{\tan }^{\frac{-2}{3}}x\Rightarrow g\left(\frac{\pi }{4}\right)=\frac{-15}{8}$