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Question Number 32928 by Cheyboy last updated on 06/Apr/18

$$\mathit{p}\mathit{l}\mathit{z}\mathit{h}\mathit{e}\mathit{l}\mathit{p}$$$$Evalute$$$$$$$$\underset{\pi /3}{\stackrel{\pi /4}{\int }}\frac{{\sin }^{2}x}{\sqrt{1-cosx}}dx$$

Commented by abdo imad last updated on 06/Apr/18

$$letputI={\int }_{\frac{\pi }{3}}^{\frac{\pi }{4}}\frac{{sin}^{2}x}{\sqrt{1-cosx}}dx$$$$I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{4}}\frac{4{sin}^2\left(\frac{x}{2}\right){cos}^2\left(\frac{x}{2}\right)}{\sqrt{2}sin\left(\frac{x}{2}\right)}dx=\frac{4}{\sqrt{2}}{\int }_{\frac{\pi }{3}}^{\frac{\pi }{4}}sin\left(\frac{x}{2}\right){cos}^2\left(\frac{x}{2}\right)dx$$$$=\frac{4}{\sqrt{2}}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}(-sin\left(\frac{x}{2}\right)){cos}^2\left(\frac{x}{2}\right)dx$$$$=\frac{4}{\sqrt{2}}\frac{2}{3}{\left[{cos}^3\left(\frac{x}{2}\right)\right]}_{\frac{\pi }{4}}^{\frac{\pi }{3}}=\frac{8}{3\sqrt{2}}({cos}^3\left(\frac{\pi }{6}\right)-{cos}^3\left(\frac{\pi }{8}\right))$$$$=\frac{8}{3\sqrt{2}}({\left(\frac{\sqrt{3}}{2}\right)}^3-{\left(\frac{\sqrt{2+\sqrt{2}}}{2}\right)}^3)$$$$=\frac{1}{3\sqrt{2}}(3\sqrt{3}-{\left(\sqrt{2+\sqrt{2}}\right)}^3).$$

Commented by Cheyboy last updated on 07/Apr/18

$$thankxyouallforurhelp$$

Answered by sma3l2996 last updated on 06/Apr/18

$$I={\int }_{\pi /3}^{\pi /4}\frac{{sin}^{2}x}{\sqrt{1-cosx}}dx$$$$lett=cosx\Rightarrow dt=-sinxdx$$$$sinx=\sqrt{1-{cos}^{2}x}=\sqrt{1-t^2}$$$$I=-{\int }_{1/2}^{\sqrt{2}/2}\frac{\sqrt{1-t^2}}{\sqrt{1-t}}dt={\int }_{\sqrt{2}/2}^{1/2}\sqrt{1+t}dt=\frac{2}{3}{\left[\sqrt{{(1+t)}^3}\right]}_{\sqrt{2}/2}^{1/2}$$$$I=\frac{2}{3}({\left(\sqrt{\frac{3}{2}}\right)}^3-{\left(\sqrt{\frac{2+\sqrt{2}}{2}}\right)}^3)$$$$etc...$$

Commented by Cheyboy last updated on 07/Apr/18

$$thankyouvrymuchsir$$

Answered by hknkrc46 last updated on 07/Apr/18

$${\int }_{\frac{\pi }{3}}^{\frac{\pi }{4}}\frac{{\sin }^2\mathrm{x}\sqrt{1-\mathrm{cosx}}}{1-\mathrm{cosx}}\mathrm{dx}={\int }_{\frac{\pi }{3}}^{\frac{\pi }{4}}\frac{\mathrm{sinx}\sqrt{1-\mathrm{cosx}}}{1-\mathrm{cosx}}\mathrm{sinxdx}$$$$={\int }_{\frac{\pi }{3}}^{\frac{\pi }{4}}\frac{\sqrt{1-{\cos }^2\mathrm{x}}\sqrt{1-\mathrm{cosx}}}{1-\mathrm{cosx}}\mathrm{sinxdx}$$$$★1-\mathrm{cosx}=\mathrm{u}\Rightarrow \mathrm{sinxdx}=\mathrm{du}\wedge (\mathrm{x}\to \frac{\pi }{4}\Rightarrow \mathrm{u}\to \frac{2-\sqrt{2}}{2}\wedge \mathrm{x}\to \frac{\pi }{3}\Rightarrow \mathrm{u}\to \frac{1}{2})$$$$={\int }_{\frac{1}{2}}^{\frac{2-\sqrt{2}}{2}}\frac{\sqrt{1-{(1-\mathrm{u})}^2}\sqrt{\mathrm{u}}}{\mathrm{u}}\mathrm{du}={\int }_{\frac{1}{2}}^{\frac{2-\sqrt{2}}{2}}\frac{\sqrt{2\mathrm{u}-{\mathrm{u}}^2}}{\sqrt{\mathrm{u}}}\mathrm{du}$$$$={\int }_{\frac{1}{2}}^{\frac{2-\sqrt{2}}{2}}\sqrt{2-\mathrm{u}}\mathrm{du}=-\frac{2}{3}{(2-\mathrm{u})}^{\frac{3}{2}}{\mid }_{\frac{1}{2}}^{\frac{2-\sqrt{2}}{2}}$$$$=-\frac{2}{3}[{(2-\frac{2-\sqrt{2}}{2})}^{\frac{3}{2}}-{(2-\frac{1}{2})}^{\frac{3}{2}}]$$$$=-\frac{2}{3}[{\left(\frac{2+\sqrt{2}}{2}\right)}^{\frac{3}{2}}-{\left(\frac{3}{2}\right)}^{\frac{3}{2}}]$$

Commented by Cheyboy last updated on 07/Apr/18

$$thankuverymuchsir$$

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