3-2-2-x-1-3-x-dx-by-substituting-x-1-cos-

Question Number 711 by prakash jain last updated on 03/Mar/15
$$\int_{\mathrm{3}/\mathrm{2}} ^{\:\mathrm{2}} \sqrt{\frac{{x}−\mathrm{1}}{\mathrm{3}−{x}}\:}{dx}\:\mathrm{by}\:\mathrm{substituting}\:{x}=\mathrm{1}−\mathrm{cos}\:\theta. \\$$
Commented by malwaan last updated on 03/Mar/15
$${I}\:{am}\:{sorry} \\$$$${x}=\mathrm{2}−{cos}\theta \\$$
Answered by prakash jain last updated on 03/Mar/15
$${x}=\mathrm{2}−\mathrm{cos}\:\theta \\$$$${dx}=\mathrm{sin}\:\theta\:{d}\theta=\mathrm{2sin}\:\frac{\theta}{\mathrm{2}}\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\$$$${x}=\mathrm{3}/\mathrm{2}\Rightarrow\theta=\pi/\mathrm{3},\:{x}=\mathrm{2}\Rightarrow\theta=\pi \\$$$$\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{3}+{x}}}=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{cos}\:\theta}}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\$$$$\int_{\pi/\mathrm{3}} ^{\:\pi/\mathrm{2}} \mathrm{2sin}\:\frac{\theta}{\mathrm{2}}\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:{d}\theta \\$$$$\int_{\pi/\mathrm{3}} ^{\:\pi/\mathrm{2}} \mathrm{2sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\:{d}\theta \\$$$$\int_{\pi/\mathrm{3}} ^{\:\pi/\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\theta\right)\:{d}\theta \\$$$$=\left[\theta−\mathrm{sin}\:\theta\right]_{\pi/\mathrm{3}} ^{\pi/\mathrm{2}} \\$$$$=\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right)−\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\$$$$=\frac{\pi}{\mathrm{6}}+\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{2}} \\$$$$\mathrm{You}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same}\:\mathrm{answer}\:\mathrm{as}\:\mathrm{in}\:\mathrm{the}\:\mathrm{earlier} \\$$$$\mathrm{question}. \\$$
Commented by malwaan1 last updated on 04/Mar/15
$${thank}\:{you} \\$$
Commented by malwaan last updated on 04/Mar/15
$${I}\:{will}\:{send}\:{my}\:{answer}\:{soon} \\$$