# Question and Answers Forum

Integration Questions

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

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

Answered by bobhans last updated on 03/Oct/20

$$\mathrm{let}\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{u}\:\mathrm{with}\:\begin{cases}{\mathrm{u}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{\mathrm{u}=\mathrm{0}}\end{cases} \\$$$$\mathrm{I}\:=\:\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}}/\mathrm{2}} {\int}}\:\frac{\mathrm{2u}\:\mathrm{du}}{\mathrm{u}^{\frac{\mathrm{4}}{\mathrm{3}}} }\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}}/\mathrm{2}} {\int}}\mathrm{u}^{−\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{du} \\$$$$\mathrm{I}=\:\mathrm{2}\left[\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{u}^{\frac{\mathrm{2}}{\mathrm{3}}} \:\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}/\mathrm{2}} =\:\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{3}\sqrt[{\mathrm{3}\:}]{\frac{\mathrm{3}}{\mathrm{4}}} \\$$$$=\:\mathrm{3}\:\sqrt[{\mathrm{3}\:}]{\frac{\mathrm{6}}{\mathrm{8}}}\:=\:\frac{\mathrm{3}\:\sqrt[{\mathrm{3}\:}]{\mathrm{6}}}{\mathrm{2}}\:. \\$$

Answered by Bird last updated on 03/Oct/20

$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{2}{sinx}\:{cosx}}{\left({sinx}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }{dx} \\$$$$=_{{sinx}={t}} \:\:\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\frac{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} }\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\$$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:{t}^{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{3}}} \:{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{t}^{−\frac{\mathrm{1}}{\mathrm{3}}} {dt} \\$$$$=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}{t}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \\$$$$=\mathrm{2}\left[\frac{\mathrm{3}}{\mathrm{2}}{t}^{\frac{\mathrm{2}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} =\mathrm{3}\left(\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \right) \\$$$$=\mathrm{3}\left(^{\mathrm{3}} \sqrt{\frac{\mathrm{3}}{\mathrm{4}}}\right) \\$$

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

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{sin2x}}{\left(\mathrm{sinx}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\mathrm{dx} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{2cosx}}{\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{x}}\mathrm{dx} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \frac{\mathrm{dt}}{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}} }=\mathrm{3}\left[\mathrm{t}^{\frac{\mathrm{2}}{\mathrm{3}}} \right]^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} =\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{3}^{\frac{\mathrm{4}}{\mathrm{3}}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \\$$