I=∫dxsin11⁡xcos⁡x3=∫dxsin113⁡cos13⁡x=∫sin−113⁡xcos−13⁡xdx Let m=−113 n=−13m+n=−113−13=−4 (is an even number) Put tan⁡x=tsec2⁡xdx=dtdx=dt1+t2∴I=∫dxsin113⁡xcos13⁡x=∫dxsin113⁡xcos113⁡x×cos113⁡xcos13⁡x=∫dxtan113⁡xcos4⁡x=∫sec2⁡xsec2⁡xtan113⁡xdx=∫1+tan2⁡xsec2⁡xtan113⁡xdx=∫1+tan2dttan113=∫t−113dt+∫t−53dt=t−83−83+t−23−23+C=−38t−83−32t−23+C=−38(tan⁡x)−83−32(tan⁡x)−23+C

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$\int \frac{d x}{\sqrt[3]{\sin^{11} x \cos x}} =$

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$\frac{- 3}{11} {(\tan x)}^{\frac{- 11}{3}} - \frac{3}{2} \tan^{\frac{- 2}{3}} x + c$

$\frac{- 3}{8} {(\tan x)}^{\frac{- 8}{3}} - \frac{3}{2} {(\tan x)}^{\frac{- 2}{3}} + c$

$\frac{- 8}{3} {(\tan x)}^{\frac{- 3}{8}} - \frac{2}{3} {(\tan x)}^{\frac{- 3}{2}} + c$

$- {(\tan x)}^{\frac{- 3}{8}} - {(\tan x)}^{\frac{- 3}{2}} + c$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{d x}{\sqrt[3]{\sin^{11} x \cos x}} = \int \frac{d x}{\sin^{\frac{11}{3}} \cos^{\frac{1}{3}} x} = \int \sin^{- \frac{11}{3}} x \cos^{- \frac{1}{3}} x d x \\ Let m = \frac{- 11}{3} n = \frac{- 1}{3} \\ m + n = \frac{- 11}{3} \frac{- 1}{3} = - 4 (is an even number) \end{array}$

$\begin{array}{l} Put \tan x = t \\ \sec^{2} x d x = d t \\ d x = \frac{d t}{1 + t^{2}} \\ ∴ I = \int \frac{d x}{\sin^{\frac{11}{3}} x \cos^{\frac{1}{3}} x} \\ = \int \frac{d x}{\frac{\sin^{\frac{11}{3}} x}{\cos^{\frac{11}{3}} x} \times \cos^{\frac{11}{3}} x \cos^{\frac{1}{3}} x} \\ = \int \frac{d x}{\tan^{\frac{11}{3}} x \cos^{4} x} \\ = \int \frac{\sec^{2} x \sec^{2} x}{\tan^{\frac{11}{3}} x} d x \\ = \int \frac{(1 + \tan^{2} x) \sec^{2} x}{\tan^{\frac{11}{3}} x} d x \end{array}$

$\begin{array}{l} = \int \frac{(1 + \tan^{2}) d t}{\tan^{\frac{11}{3}}} \\ = \int t^{\frac{- 11}{3}} d t + \int t^{\frac{- 5}{3}} d t \\ = \frac{t^{\frac{- 8}{3}}}{\frac{- 8}{3}} + \frac{t^{\frac{- 2}{3}}}{\frac{- 2}{3}} + C \\ = \frac{- 3}{8} t^{\frac{- 8}{3}} \frac{- 3}{2} t^{\frac{- 2}{3}} + C \\ = \frac{- 3}{8} (\tan x)^{\frac{- 8}{3}} \frac{- 3}{2} (\tan x)^{\frac{- 2}{3}} + C \end{array}$