If \(\int \frac{\mathrm{d} x}{\sqrt[3]{\sin ^{11} x \cos x}}=-\left(\frac{3}{8} \mathrm{f}(x)+\frac{3}{2} \mathrm{~g}(x)\right)+\mathrm{c}\) then

Let
\(\mathrm{I} =\int \frac{\mathrm{d} x}{\sqrt[3]{\sin ^{11} x \cdot \cos x}} \)
\( =\int \frac{\mathrm{d} x}{\sin ^{\frac{11}{3}} x \cdot \cos ^{\frac{1}{3}} x} \)
\( =\int \frac{\sec ^4 x}{\tan ^{\frac{11}{3}} x} \mathrm{~d} x\)
... [Dividing numerator and denominator by \(\left.\cos ^{\frac{11}{3}} x\right]\)
\(=\int \frac{\left(1+\tan ^2 x\right) \sec ^2 x}{\tan ^{\frac{11}{3}} x} \mathrm{~d} x\)
\(\text {Let } \tan x=t \)
\( \sec ^2 x d x=d t \)
\( \therefore =\int \frac{\left(1+t^2\right) d t}{t^{\frac{11}{3}}} \)
\( =\int t^{-\frac{11}{3}} d t+\int t^{\frac{-5}{3}} d t\)
\(=\frac{-3}{8} \mathrm{t}^{\frac{-8}{3}}-\frac{3}{2} \mathrm{t}^{\frac{-2}{3}}+\mathrm{c} \)
\( =-\left[\frac{3}{8} \mathrm{t}^{\frac{-8}{3}}+\frac{3}{2} \mathrm{t}^{\frac{-2}{3}}\right]+\mathrm{c} \)
\( =-\left[\frac{3}{8} \tan ^{\frac{-8}{3}} x+\frac{3}{2} \tan ^{\frac{-2}{3}} x\right]+\mathrm{c} \)
\( \therefore \mathrm{f}(x) =\tan ^{\frac{-8}{3}} x, \mathrm{~g}(x)=\tan ^{\frac{-2}{3}} x\)