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Question Number 140930 by bramlexs22 last updated on 14/May/21

$$\underset{0}{\stackrel{\pi }{\int }}\frac{\mathrm{dx}}{\sqrt{2}-\cos \mathrm{x}}$$

Answered by Dwaipayan Shikari last updated on 14/May/21

$$2{\int }_0^{\mathrm{\infty }}\frac{dt}{\sqrt{2}-\frac{1-t^2}{1+t^2}}.\frac{1}{1+t^2}t=tan\frac{x}{2}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dt}{\sqrt{2}{t}^2+\sqrt{2}-1+t^2}dt=\frac{1}{\sqrt{2}+1}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dt}{t^2+\frac{\sqrt{2}-1}{\sqrt{2}+1}}dt$$$${=\frac{1}{\sqrt{2}+1}.\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}[}{tan}^{-1}\left(t\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}\right)]}_{-\mathrm{\infty }}^{\mathrm{\infty }}dx$$$$=\pi$$

Commented by bramlexs22 last updated on 14/May/21

$$\mathrm{Weirrstrass}\mathrm{substitution}$$

Commented by Dwaipayan Shikari last updated on 14/May/21

$$yes!$$

Answered by mathmax by abdo last updated on 14/May/21

$$\mathrm{I}={\int }_0^{\pi }\frac{\mathrm{dx}}{\sqrt{2}-\mathrm{cosx}}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}\Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(\sqrt{2}-\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{\sqrt{2}+\sqrt{2}{\mathrm{t}}^2-1+{\mathrm{t}}^2}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{(1+\sqrt{2}){\mathrm{t}}^2+\sqrt{2}-1}=\frac{2}{1+\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\frac{\sqrt{2}-1}{\sqrt{2}+1}}$$$$=\frac{2}{1+\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+3-2\sqrt{2}}{=}_{\mathrm{t}=\sqrt{3-2\sqrt{2}}\mathrm{u}}\frac{2}{1+\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{3-2\sqrt{2}}\mathrm{du}}{(3-2\sqrt{2})(1+{\mathrm{u}}^2)}$$$$=\frac{2}{(1+\sqrt{2})\sqrt{3-2\sqrt{2}}}{\left[\mathrm{arctanu}\right]}_0^{\mathrm{\infty }}=\frac{\pi }{(1+\sqrt{2})\sqrt{3-2\sqrt{2}}}$$

Commented by mathmax by abdo last updated on 14/May/21

$$\mathrm{but}\sqrt{3-2\sqrt{2}}=\sqrt{2}-1\Rightarrow \mathrm{I}=\frac{\pi }{(\sqrt{2}-1)(\sqrt{2}+1)}=\pi$$$$$$

Answered by iloveisrael last updated on 14/May/21

$$R=\underset{0}{\stackrel{\pi }{\int }}\frac{dx}{a-\cos x};a>1$$$$put\mu =\tan \frac{x}{2}$$$$R=\underset{0}{\stackrel{\pi }{\int }}\frac{1+{\mu }^2}{a+a{\mu }^2-1+{\mu }^2}.\frac{2}{1+{\mu }^2}d\mu$$$$(\ast )a-1+(1+a){\mu }^2;set\frac{a-1}{a+1}=p^2$$$$and\mu =pu$$$$R=\frac{2}{a+1}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{p}{p^2(1+u^2)}du$$$$R=\frac{2}{\sqrt{a^2-1}}\left[\arctan u\right]{}_0^{\mathrm{\infty }}$$$$R=\frac{2}{\sqrt{a^2-1}}.\frac{\pi }{2}=\frac{\pi }{\sqrt{a^2-1}}$$$$R=\frac{\pi }{\sqrt{{\left(\sqrt{2}\right)}^2-1}}=\pi$$

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