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Question Number 33759 by 33 last updated on 24/Apr/18

$$solve:$$$$\underset{-\pi /2}{\stackrel{\pi /2}{\int }}\frac{sin\theta }{\sqrt{R^2+r^2-2rRcos\theta }}d\theta$$

Commented by MJS last updated on 24/Apr/18

$$\int \frac{\cos \theta }{\sqrt{R^2+r^2-2rR\cos \theta }}d\theta =$$$$R^2+r^2=p$$$$-2rR=q$$$$=\int \frac{\cos \theta }{\sqrt{p+q\cos \theta }}d\theta =$$$$\mathrm{thanks}\mathrm{to}\mathrm{my}\mathrm{friend}\mathrm{Prof}.\mathrm{H}.\mathrm{E}.$$$$\cos \theta =\frac{1}{q}(p+q\cos \theta )-\frac{p}{q}$$$$=\int \frac{\frac{1}{q}(p+q\cos \theta )-\frac{p}{q}}{\sqrt{p+q\cos \theta }}d\theta =$$$$=\frac{1}{q}\int \frac{p+q\cos \theta }{\sqrt{p+q\cos \theta }}d\theta -\frac{p}{q}\int \frac{1}{\sqrt{p+q\cos \theta }}d\theta =$$$$=\frac{1}{q}\int \sqrt{p+q\cos \theta }d\theta -\frac{p}{q}\int \frac{1}{\sqrt{p+q\cos \theta }}d\theta =$$$${\sin }^2\frac{\theta }{2}=\frac{1-\cos \theta }{2}$$$$\cos \theta =1-{2\sin }^2\frac{\theta }{2}$$$$p+q\cos \theta =p+q-2q{\sin }^2\frac{\theta }{2}=$$$$=(p+q)(1-\frac{2q}{p+q}{\sin }^2\frac{\theta }{2})$$$$=\frac{\sqrt{p+q}}{q}\int \sqrt{(1-\frac{2q}{p+q}{\sin }^2\frac{\theta }{2})}d\theta -\frac{p\sqrt{p+q}}{q(p+q)}\int \frac{1}{\sqrt{(1-\frac{2q}{p+q}{\sin }^2\frac{\theta }{2})}}d\theta$$$$u=\frac{x}{2}\to dx=2du$$$$=\frac{2\sqrt{p+q}}{q}\int \sqrt{(1-\frac{2q}{p+q}{\sin }^{2}u)}du-\frac{2p\sqrt{p+q}}{q(p+q)}\int \frac{1}{\sqrt{(1-\frac{2q}{p+q}{\sin }^{2}u)}}du=$$$$\mathrm{the}\mathrm{first}\mathrm{is}\mathrm{an}\mathrm{elliptic}\mathrm{integral}\mathrm{of}\mathrm{the}$$$$2^{\mathrm{nd}}\mathrm{kind},\mathrm{the}\mathrm{second}\mathrm{one}\mathrm{of}\mathrm{the}{1}^{\mathrm{st}}\mathrm{kind}$$$$\mathrm{the}\mathrm{notation}\mathrm{follows},\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{beyond}$$$$\mathrm{my}\mathrm{understanding}...$$$$=\frac{2\sqrt{\mathrm{p}+\mathrm{q}}}{\mathrm{q}}E(u\mid \frac{2q}{p+q})-\frac{2p\sqrt{p+q}}{q(p+q)}F(u\mid \frac{2q}{p+q})=$$$$u=\frac{x}{2}$$$$=\frac{2\sqrt{p+q}}{q}(E(\frac{x}{2}\mid \frac{2q}{p+q})-\frac{p}{p+q}F(\frac{x}{2}\mid \frac{2q}{p+q}))$$

Commented by prof Abdo imad last updated on 24/Apr/18

$$letputI={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{cos\theta }{\sqrt{R^2+r^2-2rRcos\theta }}d\theta$$$$wehave{R}^2+r^2-2rRcos\theta$$$$=R^2(1+{\left(\frac{r}{R}\right)}^2-2\frac{r}{R}cos\theta letput\lambda =\frac{r}{R}$$$$I=\frac{1}{R}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{cos\theta }{\sqrt{1+{\lambda }^2-2\lambda cos\theta }}d\theta$$$$tan\left(\frac{\theta }{2}\right)=t\Rightarrow I=\frac{1}{R}{\int }_{-1}^1\frac{\frac{1-t^2}{1+t^2}}{\sqrt{1+{\lambda }^2-2\lambda \frac{1-t^2}{1+t^2}}}\frac{2dt}{1+t^2}$$$$I=\frac{2}{R}{\int }_{-1}^1\frac{1-t^2}{{(1+t^2)}^2}\sqrt{1+t^2}\frac{dt}{\sqrt{(1+{\lambda }^2)(1+t^2)-2\lambda (1-t^2)}}$$$$=\frac{2}{R}{\int }_{-1}^1\frac{1-t^2}{{(1+t^2)}^{\frac{3}{2}}}\frac{dt}{\sqrt{1+t^2+{\lambda }^2+{\lambda }^2{t}^2-2\lambda +2\lambda t^2}}$$$$=\frac{2}{R}{\int }_{-1}^1\frac{(1-t^2)dt}{{(1+t^2)}^{\frac{3}{2}}\sqrt{{\lambda }^2-2\lambda +1+({\lambda }^2+2\lambda +1)t^2}}$$$$=\frac{2}{R}{\int }_{-1}^1\frac{1-t^2}{{(1+t^2)}^{\frac{3}{2}}\sqrt{{(\lambda -1)}^2+{(\lambda +1)}^2{t}^2}}dt$$$$ch.(\lambda +1)t=(\lambda -1)ugive$$$$I=\frac{2}{R}{\int }_{-1}^1\frac{1-{\left(\frac{\lambda -1}{\lambda +1}u\right)}^2}{{(1+{\left(\frac{\lambda -1}{\lambda +1}u\right)}^2)}^{\frac{3}{2}}}\frac{\frac{\lambda -1}{\lambda +1}}{\mid \lambda -1\mid \sqrt{1+u^2}}du$$$$afterweusethech.u=tan\theta oru=sh\theta ...be$$$$continued...$$

Commented by 33 last updated on 24/Apr/18

$$TosirMJS,sirwecannot$$$$considera,bandcas$$$$constantsastheyare$$$$dependentoncos\theta .$$$$andyoudonotknow$$$$abcasafunctionof\alpha$$$$andviceversa.$$

Commented by MJS last updated on 24/Apr/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}$$

Commented by MJS last updated on 24/Apr/18

$$...\mathrm{this}\mathrm{leads}\mathrm{to}\mathrm{an}\mathrm{elliptic}\mathrm{integral}\mathrm{and}\mathrm{it}$$$$\mathrm{might}\mathrm{be}\mathrm{impossible}\mathrm{to}\mathrm{exactly}\mathrm{solve}\mathrm{it}$$

Commented by 33 last updated on 24/Apr/18

$$whatifthereweresine$$$$insteadofcosinnumerator?$$

Commented by MJS last updated on 24/Apr/18

$$\mathrm{then}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}$$$$\int \frac{\cos \theta }{\sqrt{R^2+r^2-2Rr\sin \theta }}d\theta =$$$$u=R^2+r^2-2Rr\sin \theta$$$$dx=-\frac{1}{2rR\cos \theta }$$$$-\frac{1}{2rR}\int \frac{1}{\sqrt{u}}du=-\frac{1}{2rR}\times 2\sqrt{u}+C=-\frac{\sqrt{u}}{rR}+C=$$$$=-\frac{\sqrt{R^2+r^2-2Rr\sin \theta }}{rR}+C$$$$\underset{-\frac{\pi }{2}}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\cos \theta }{\sqrt{R^2+r^2-2Rr\sin \theta }}d\theta =$$$$=\frac{\sqrt{R^2+r^2+2rR}-\sqrt{R^2+r^2-2rR}}{rR}=$$$$=\frac{\sqrt{{(R+r)}^2}-\sqrt{{(R-r)}^2}}{rR}=$$$$=\frac{\mid R+r\mid -\mid R-r\mid }{rR}=$$$$\mathrm{if}R>r\wedge R>0\wedge r>0$$$$=\frac{R+r-(R-r)}{rR}=\frac{2}{R}$$

Commented by MJS last updated on 24/Apr/18

$$\mathrm{sorry},\mathrm{I}\mathrm{misread}...\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}\mathrm{way}:$$$$\int \frac{\sin \theta }{\sqrt{R^2+r^2-2Rr\cos \theta }}d\theta =$$$$=\frac{\sqrt{R^2+r^2-2Rr\cos \theta }}{rR}+C$$$$\underset{-\frac{\pi }{2}}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\sin \theta }{\sqrt{R^2+r^2-2Rr\cos \theta }}d\theta =0$$

Commented by prof Abdo imad last updated on 25/Apr/18

$$somethingwentwrongbecsuse\theta \to \frac{sin\theta }{\sqrt{R^2+r^2-2rRcos\theta }}$$$$isodd\Rightarrow {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(...)d\theta =0$$

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