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We have to evaluate the following integration: $$ \int(\sin x)^{-11/3}(\cos x)^{-1/3}\,dx $$

For this I tried.

And I'm uncertain if we can use Wallis’ theorem.

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Hint write it as $\frac {\sec^4 (x)}{(\tan (x))^{11/3}} $ put $\tan(x)=t $ but $sec^2 (x)dx=dt $ then integral reduces to $\frac {1+t^2}{t^{11/3}} $ .Now split the terms and its easily integrable.

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$$f(x)=$$ $(\sin(x))^{\frac{ -11 }{ 3 }}(1-(\sin(x))^2)^{\frac{-2}{3}}\cos(x) dx$

put $\sin(x)=t$. then the integral becomes $$I=-\int t^{\frac{-11}{3}}(1-t^2)^{\frac{-2}{3}} dt$$

now put $u=t^{\frac{-8}{3}}$ and you can finish.

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Divide numerator and denominator by cos^x to convert the given function in terms of sec^2x and tanx Then substitute tanx=t and proceed

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    This would be easier to read if you had used MathJax to format it. There's no clear numerator/denominator in the question, making it harder to know what you mean, and you've answered a question that is a little over a month old and has an accepted answer(which says something similar to yours). That is three reasons this is not really a worthwhile addition to the site, or a good use of your time. – 2018-02-15