Question

Integrate cotx/cuberoot of sinx dx

Answer 1

$\underline{\textsf{To find:}}$

$\mathsf{\displaystyle\int\,\dfrac{cotx}{\sqrt[3]{sinx}}\,dx}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\displaystyle\int\,\dfrac{cotx}{\sqrt[3]{sinx}}\,dx}$

$\mathsf{=\displaystyle\int\,\dfrac{cosx}{sinx}\dfrac{1}{\sqrt[3]{sinx}}\,dx}$

$\mathsf{=\displaystyle\int\,\dfrac{cosx}{sinx}\dfrac{1}{(sinx)^\frac{1}{3}}\,dx}$

$\mathsf{=\displaystyle\int\,\dfrac{cosx}{(sinx)^\frac{4}{3}}\,dx}$

$\boxed{\begin{minipage}{3cm} \\\textsf{Take}\,\mathsf{t=sinx}\\\\\mathsf{\dfrac{dt}{dx}=cosx}\\\\\mathsf{dt=cosx\,dx} \end{minipage}}$

$\mathsf{=\displaystyle\int\,\dfrac{dt}{t^\frac{4}{3}}}$

$\mathsf{=\displaystyle\int\,t^\frac{-4}{3}\,dt}$

$\mathsf{=\dfrac{t^{\frac{-4}{3}+1}}{\frac{-4}{3}+1}+C}$

$\mathsf{=\dfrac{t^\frac{-1}{3}}{\frac{-1}{3}}+C}$

$\mathsf{=\dfrac{-3}{t^\frac{1}{3}}+C}$

$\mathsf{=\dfrac{-3}{(sinx)^\frac{1}{3}}+C}$

$\underline{\textsf{Answer:}}$

$\mathsf{\displaystyle\int\,\dfrac{cotx}{\sqrt[3]{sinx}}\,dx=\dfrac{-3}{(sinx)^\frac{1}{3}}+C}$

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