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Question Number 129105 by I want to learn more last updated on 12/Jan/21

$${\int }_0^{\frac{\pi }{4}}\sqrt{\tan \mathrm{x}(1-\tan \mathrm{x})}\mathrm{dx}$$

Answered by Lordose last updated on 13/Jan/21

$$$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}\sqrt{\tan \left(\mathrm{x}\right)}\sqrt{1-\tan \left(\mathrm{x}\right)}\mathrm{dx}\stackrel{\mathrm{u}=\tan \left(\mathrm{x}\right)}{=}{\int }_0^1\frac{{\mathrm{u}}^{\frac{1}{2}}\sqrt{1-\mathrm{u}}}{1+{\mathrm{u}}^2}\mathrm{du}$$$$\mathrm{\Omega }={\int }_0^1{\mathrm{u}}^{\frac{1}{2}}{(1-\mathrm{u})}^{\frac{1}{2}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{2\mathrm{n}}\mathrm{du}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{u}}^{2\mathrm{n}+\frac{1}{2}}{(1-\mathrm{u})}^{\frac{1}{2}}\mathrm{du}$$$$\mathit{\beta }(\mathrm{x},\mathrm{y})={\int }_0^1{\mathrm{t}}^{\mathrm{x}-1}{(1-\mathrm{t})}^{\mathrm{y}-1}\mathrm{dt}$$$$\mathrm{\Omega }=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\left(\mathit{\beta }(\frac{4\mathrm{n}+3}{2},\frac{3}{2})\right)=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\left(\frac{\mathbf{\Gamma }\left(\frac{4\mathrm{n}+3}{2}\right)\mathbf{\Gamma }\left(\frac{3}{2}\right)}{\mathbf{\Gamma }(2\mathrm{n}+3)}\right)$$$$\mathrm{\Omega }=\frac{\sqrt{\pi }}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\left(\frac{\mathbf{\Gamma }\left(\frac{4\mathrm{n}+3}{2}\right)}{\mathbf{\Gamma }(2\mathrm{n}+3)}\right)=\frac{\sqrt{\pi }}{2}(\sqrt{\pi }(\sqrt{2(1+\sqrt{2})}-2)=\frac{\pi }{2}(\sqrt{2(1+\sqrt{2})}-2)$$$$★\mathbf{L}\mathit{\varphi }\mathbf{rD}\mathit{\varnothing }\mathbf{sE}$$

Commented by MJS_new last updated on 13/Jan/21

$$\mathrm{great}!$$

Commented by I want to learn more last updated on 13/Jan/21

$$\mathrm{Thanks}\mathrm{sir}.$$

Answered by MJS_new last updated on 13/Jan/21

$$\int \sqrt{(1-\tan x)\tan x}dx=$$$$[t=\sqrt{\frac{1}{\tan x}-1}\to dx=-\frac{2\sqrt{(1-\tan x){\tan }^{3}x}}{1+{\tan }^{2}x}dt]$$$$=-2\int \frac{t^2}{(t^2+1)(t^4+2t^2+2)}dt=$$$$=-2\int \frac{t^2}{(t^2+1)(t^2-\sqrt{-2+2\sqrt{2}}t+\sqrt{2})(t^2+\sqrt{-2+2\sqrt{2}}t+\sqrt{2})}dt=$$$$=-2\int \frac{t^2}{(t^2+1)(t^2-at+b)(t^2+at+b)}dt=$$$$=2\int \frac{dt}{t^2+1}+\frac{\sqrt{-2+2\sqrt{2}}}{2}(\int \frac{t-2\sqrt{1+\sqrt{2}}}{t^2-\sqrt{-2+2\sqrt{2}}t+\sqrt{2}}dt-\int \frac{t+2\sqrt{1+\sqrt{2}}}{t^2+\sqrt{-2+2\sqrt{2}}t+\sqrt{2}}dt)$$$$\mathrm{and}\mathrm{these}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{using}\mathrm{the}\mathrm{usual}\mathrm{formulas}$$$$\mathrm{in}\mathrm{the}\mathrm{end}\mathrm{I}\mathrm{get}$$$$(\frac{\sqrt{2+2\sqrt{2}}}{2}-1)\pi$$

Commented by I want to learn more last updated on 13/Jan/21

$$\mathrm{Thanks}\mathrm{sir}.$$

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