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

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Let $\begin{aligned} \mathrm{I} & =\int \frac{\mathrm{d} x}{\sqrt[3]{\sin ^{11} x \cdot \cos x}} \ & =\int \frac{\mathrm{d} x}{\sin ^{\frac{11}{3}} x \cdot \cos ^{\frac{1}{3}} x} \ & =\int \frac{\sec ^4 x}{	an ^{\frac{11}{3}} x} \mathrm{~d} x\end{aligned}$ ... [Dividing numerator and denominator by $\left.\cos ^{\frac{11}{3}} xight]$ $=\int \frac{\left(1+	an ^2 xight) \sec ^2 x}{	an ^{\frac{11}{3}} x} \mathrm{~d} x$ $\begin{aligned} & 	ext { Let } 	an x=t \ & \sec ^2 x d x=d t \ 	herefore \quad & =\int \frac{\left(1+t^2ight) d t}{t^{\frac{11}{3}}} \ & =\int t^{-\frac{11}{3}} d t+\int t^{\frac{-5}{3}} d t\end{aligned}$ $\begin{aligned} & =\frac{-3}{8} \mathrm{t}^{\frac{-8}{3}}-\frac{3}{2} \mathrm{t}^{\frac{-2}{3}}+\mathrm{c} \ & =-\left[\frac{3}{8} \mathrm{t}^{\frac{-8}{3}}+\frac{3}{2} \mathrm{t}^{\frac{-2}{3}}ight]+\mathrm{c} \ & =-\left[\frac{3}{8} 	an ^{\frac{-8}{3}} x+\frac{3}{2} 	an ^{\frac{-2}{3}} xight]+\mathrm{c} \ 	herefore \quad \mathrm{f}(x) & =	an ^{\frac{-8}{3}} x, \mathrm{~g}(x)=	an ^{\frac{-2}{3}} x\end{aligned}$

If $\int \frac{\mathrm{d} x}{\sqrt[3]{\sin ^{11} x \cos x}}=-\left(\frac{3}{8} \mathrm{f}(x)+\frac{3}{2}…

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