Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}{\sec }^{\frac{2}{3}}x·{\mathrm{cosec}}^{\frac{4}{3}}xdx$

$\Rightarrow I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\left(\frac{1}{{\cos }^{\frac{2}{3}}x·{\sin }^{\frac{4}{3}}x}\right)dx$

$\Rightarrow I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\left(\frac{{\cos }^{{\displaystyle \frac{4}{3}}}x}{{\cos }^{\frac{2}{3}+{\displaystyle \frac{4}{3}}}x·{\sin }^{{\displaystyle \frac{4}{3}}}x}\right)dx$

$\Rightarrow I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\left(\frac{{\sec }^{2}x}{{\tan }^{{\displaystyle \frac{4}{3}}}x}\right)dx$

Let $\tan x=t,{\mathrm{ sec}}^{2}xdx=dt$ and at $x=\frac{\mathrm{\pi }}{6}, t=\tan \left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{\sqrt{3}}$ and at $x=\frac{\mathrm{\pi }}{3}, t=\tan \left(\frac{\mathrm{\pi }}{3}\right)=\sqrt{3}$

$\Rightarrow I={\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{dt}{t^{\frac{4}{3}}}$

Using $\int x^{n}dx=\frac{x^{n+1}}{n+1}$

$\Rightarrow I={\left[\frac{t^{-\frac{4}{3}+1}}{-\frac{4}{3}+1}\right]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$

$\Rightarrow I={\left[-3t^{-\frac{1}{3}}\right]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$

$\Rightarrow I=-3\left\{{\left(\sqrt{3}\right)}^{-\frac{1}{3}}-{\left(\frac{1}{\sqrt{3}}\right)}^{-\frac{1}{3}}\right\}$

$\Rightarrow I=-3\left\{{\left(3^{\frac{1}{2}}\right)}^{-\frac{1}{3}}-{\left(\frac{1}{3^{\frac{1}{2}}}\right)}^{-\frac{1}{3}}\right\}$

$=-3\left(\frac{1}{3^{\frac{1}{6}}}-3^{\frac{1}{6}}\right)$

$=\frac{-3}{3^{\frac{1}{6}}}+3\times 3^{\frac{1}{6}}$

${=3}^{\frac{7}{6}}-3^{\frac{5}{6}}.$