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Question Number 116311 by bemath last updated on 03/Oct/20

$$\underset{0}{\stackrel{\frac{\pi }{3}}{\int }}\frac{\sin 2\mathrm{x}}{{\left(\sin \mathrm{x}\right)}^{\frac{4}{3}}}\mathrm{dx}$$$$$$

Answered by bobhans last updated on 03/Oct/20

$$\mathrm{let}\sin \mathrm{x}=\mathrm{u}\mathrm{with}\{\begin{array}{l}\mathrm{u}=\frac{\sqrt{3}}{2}\\ \mathrm{u}=0\end{array}$$$$\mathrm{I}=\underset{0}{\stackrel{\sqrt{3}/2}{\int }}\frac{2\mathrm{u}\mathrm{du}}{{\mathrm{u}}^{\frac{4}{3}}}=2\underset{0}{\stackrel{\sqrt{3}/2}{\int }}{\mathrm{u}}^{-\frac{1}{3}}\mathrm{du}$$$$\mathrm{I}=2{\left[\frac{3}{2}{\mathrm{u}}^{\frac{2}{3}}\right]}_0^{\sqrt{3}/2}=3{\left(\frac{\sqrt{3}}{2}\right)}^{\frac{2}{3}}=3\sqrt[3]{\frac{3}{4}}$$$$=3\sqrt[3]{\frac{6}{8}}=\frac{3\sqrt[3]{6}}{2}.$$

Answered by Bird last updated on 03/Oct/20

$$I={\int }_0^{\frac{\pi }{3}}\frac{2sinxcosx}{{\left(sinx\right)}^{\frac{4}{3}}}dx$$$${=}_{sinx=t}2{\int }_0^{\frac{\sqrt{3}}{2}}\frac{t\sqrt{1-t^2}}{t^{\frac{4}{3}}}\frac{dt}{\sqrt{1-t^2}}$$$$=2{\int }_0^{\frac{\sqrt{3}}{2}}{t}^{1-\frac{4}{3}}dt=2{\int }_0^{\frac{\sqrt{3}}{2}}{t}^{-\frac{1}{3}}dt$$$$=2{\left[\frac{1}{1-\frac{1}{3}}{t}^{1-\frac{1}{3}}\right]}_0^{\frac{\sqrt{3}}{2}}$$$$=2{\left[\frac{3}{2}{t}^{\frac{2}{3}}\right]}_0^{\frac{\sqrt{3}}{2}}=3\left({\left(\frac{\sqrt{3}}{2}\right)}^{\frac{2}{3}}\right)$$$$=3\left({}^3\sqrt{\frac{3}{4}}\right)$$

Answered by Dwaipayan Shikari last updated on 03/Oct/20

$${\int }_0^{\frac{\pi }{3}}\frac{\sin 2\mathrm{x}}{{\left(\mathrm{sinx}\right)}^{\frac{4}{3}}}\mathrm{dx}$$$${\int }_0^{\frac{\pi }{3}}\frac{2\mathrm{cosx}}{{\sin }^{\frac{1}{3}}\mathrm{x}}\mathrm{dx}$$$$=2{\int }_0^{\frac{\sqrt{3}}{2}}\frac{\mathrm{dt}}{{\mathrm{t}}^{\frac{1}{3}}}=3{\left[{\mathrm{t}}^{\frac{2}{3}}\right]}^{\frac{\sqrt{3}}{2}}=3{\left(\frac{\sqrt{3}}{2}\right)}^{\frac{2}{3}}=3^{\frac{4}{3}}{\left(\frac{1}{2}\right)}^{\frac{2}{3}}$$

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