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Question Number 32810 by mondodotto@gmail.com last updated on 02/Apr/18

Commented by caravan msup abdo. last updated on 02/Apr/18

$$letusethech.\sqrt{3}x=2sint$$$$I={\int }_0^{\frac{\pi }{3}}\frac{2}{\sqrt{3}}sint\sqrt{4-4{sin}^{2}t}\frac{2}{\sqrt{3}}costdt$$$$I=\frac{8}{3}{\int }_0^{\frac{\pi }{3}}sint{cos}^{2}tdt$$$$=\frac{8}{3}{[-\frac{1}{3}{cos}^{3}t]}_0^{\frac{\pi }{3}}$$$$=-\frac{8}{9}({\left(\frac{1}{2}\right)}^3-1)$$$$=-\frac{8}{9}(\frac{1}{8}-1)=-\frac{8}{9}(-\frac{7}{8})\Rightarrow$$

Commented by caravan msup abdo. last updated on 02/Apr/18

$$I=\frac{7}{9}.$$

Answered by Joel578 last updated on 02/Apr/18

$$u=4-3x^2\to du=-6xdx$$$$I=-\frac{1}{6}{\int }_4^1\sqrt{u}du$$$$=-\frac{1}{6}{\left[\frac{2}{3}u\sqrt{u}\right]}_4^1$$$$=-\frac{1}{6}(\frac{2}{3}\sqrt{1}-\frac{8}{3}\sqrt{4})$$

Commented by mondodotto@gmail.com last updated on 02/Apr/18

$$\mathrm{how}\mathrm{come}\mathrm{this}{\int }_4^1???$$

Commented by Joel578 last updated on 02/Apr/18

$$\mathrm{you}\mathrm{need}\mathrm{to}\mathrm{change}\mathrm{because}\mathrm{the}\mathrm{substitution}u$$$$u=4-3x^2$$$$\mathrm{Hence},\mathrm{when}x=0\to u=4$$$$x=1\to u=1$$

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