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Question Number 122631 by MJS_new last updated on 18/Nov/20

$$\mathrm{solve}\underset{{(a-1)}^2}{\stackrel{a^2}{\int }}{\cosh }^{-1}\frac{1}{\sqrt{a-\sqrt{x}}}dx\mathrm{with}a>0$$

Commented byliberty last updated on 18/Nov/20

$$waw..nicequestion$$

Commented byMJS_new last updated on 19/Nov/20

$$f\left(a\right)=\underset{{(a-1)}^2}{\stackrel{a^2}{\int }}{\cosh }^{-1}\frac{1}{\sqrt{a-\sqrt{x}}}dx\mathrm{is}\mathrm{linear}\mathrm{for}a⩾1$$ $$\mathrm{I}\mathrm{found}\mathrm{a}\mathrm{nice}\mathrm{path}\mathrm{to}\mathrm{solve}\mathrm{it},\mathrm{will}\mathrm{post}\mathrm{it}\mathrm{later}$$

Answered by mathmax by abdo last updated on 18/Nov/20

$$\mathrm{I}={\int }_{{(\mathrm{a}-1)}^2}^{{\mathrm{a}}^2}\mathrm{argch}\left(\frac{1}{\sqrt{\mathrm{a}-\sqrt{\mathrm{x}}}}\right)\mathrm{dx}\mathrm{chamgement}\sqrt{\mathrm{a}-\sqrt{\mathrm{x}}}=\mathrm{t}\mathrm{give}$$ $$\mathrm{a}+\sqrt{\mathrm{x}}={\mathrm{t}}^2\Rightarrow \sqrt{\mathrm{x}}={\mathrm{t}}^2-\mathrm{a}\Rightarrow \mathrm{x}={({\mathrm{t}}^2-\mathrm{a})}^2\Rightarrow$$ $$\mathrm{I}={\int }_1^{\mathrm{o}}\mathrm{argch}\left(\frac{1}{\mathrm{t}}\right)4\mathrm{t}({\mathrm{t}}^2-\mathrm{a})\mathrm{dt}=-4{\int }_{\mathrm{o}}^1\mathrm{t}({\mathrm{t}}^2-\mathrm{a})\ln (\frac{1}{\mathrm{t}}+\sqrt{\frac{1}{{\mathrm{t}}^2}-1})\mathrm{dt}$$ $$=-4{\int }_0^1\mathrm{t}({\mathrm{t}}^2-\mathrm{a})\ln \left(\frac{1+\sqrt{1-{\mathrm{t}}^2}}{\mathrm{t}}\right)\mathrm{dt}$$ $${=}_{\mathrm{t}=\sin \theta }-4{\int }_0^{\frac{\pi }{2}}\sin \theta ({\sin }^2\theta -\mathrm{a})\ln \left(\frac{1+\cos \theta }{\sin \theta }\right)\cos \theta \mathrm{d}\theta$$ $$=-4{\int }_0^{\frac{\pi }{2}}\sin \theta ({\sin }^2\theta -\mathrm{a})\ln \left(\frac{{2\cos }^2\left(\frac{\theta }{2}\right)}{2\cos \left(\frac{\theta }{2}\right)\sin \left(\frac{\theta }{2}\right)}\right)\cos \theta \mathrm{d}\theta$$ $$=4{\int }_0^{\frac{\pi }{2}}\sin \theta \cos \theta ({\sin }^2\theta -\mathrm{a})\ln \left(\tan \left(\frac{\theta }{2}\right)\right)\mathrm{d}\theta$$ $$=4{\int }_0^{\frac{\pi }{2}}({\sin }^3\theta \cos \theta -\sin \theta \cos \theta )\ln \left(\tan \left(\frac{\theta }{2}\right)\right)\mathrm{d}\theta \mathrm{this}\mathrm{integral}\mathrm{can}\mathrm{be}$$ $$\mathrm{solved}\mathrm{by}\mathrm{parts}\mathrm{if}\mathrm{we}\mathrm{put}{\mathrm{u}}^{{}^{\prime }}={\sin }^3\cos \theta -\sin \theta \cos \theta \mathrm{and}\mathrm{v}=\ln \left(\tan \left(\frac{\theta }{2}\right)\right)$$ $$...\mathrm{be}\mathrm{continued}...$$

Commented byMJS_new last updated on 19/Nov/20

$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{trying}$$

Answered by MJS_new last updated on 19/Nov/20

$$\int {\cosh }^{-1}\frac{1}{\sqrt{a-\sqrt{x}}}dx=$$ $$\mathrm{by}\mathrm{parts}\mathrm{but}\mathrm{with}\mathrm{a}\mathrm{simple}\mathrm{trick}:$$ $$u^{{}^{\prime }}=1\to u=x-a^2$$ $$v={\cosh }^{-1}\frac{1}{\sqrt{a-\sqrt{x}}}\to v^{\prime }=-\frac{1}{4\sqrt{x}(\sqrt{x}-a)\sqrt{\sqrt{x}-a+1}}$$ $$=(x-a^2){\cosh }^{-1}\frac{1}{\sqrt{a-\sqrt{x}}}+\frac{1}{4}\int \frac{x-a^2}{\sqrt{x}(\sqrt{x}-a)\sqrt{\sqrt{x}-a+1}}dx=$$ $$$$ $$\frac{1}{4}\int \frac{x-a^2}{\sqrt{x}(\sqrt{x}-a)\sqrt{\sqrt{x}-a+1}}dx=$$ $$[t=\sqrt{\sqrt{x}-a+1}\to dx=4\sqrt{x}\sqrt{\sqrt{x}-a+1}dt]$$ $$=\int (t^2+2a-1)dt=\frac{1}{3}t(t^2+3(2a-1))=$$ $$=\frac{1}{3}(\sqrt{x}+5a-2)\sqrt{\sqrt{x}-a+1}$$ $$$$ $$=(x-a^2){\cosh }^{-1}\frac{1}{\sqrt{a-\sqrt{x}}}+\frac{1}{3}(\sqrt{x}+5a-2)\sqrt{\sqrt{x}-a+1}+C$$ $$$$ $$\mathrm{inserting}\mathrm{the}\mathrm{borders}\mathrm{we}\mathrm{get}$$ $$f\left(a\right)=(2a-1){\cosh }^{-1}\frac{1}{\sqrt{a-\mid a-1\mid }}-\frac{(5a+\mid a-1\mid -2)\sqrt{1-a+\mid a-1\mid }}{3}+\frac{(5a+\mid a\mid -2)\sqrt{1-a+\mid a\mid }}{3}$$ $$\mathrm{for}0<a<1\mathrm{we}\mathrm{get}$$ $$f\left(a\right)=(2a-1){\cosh }^{-1}\frac{1}{\sqrt{2a-1}}-\frac{(4a-1)\sqrt{2(1-a)}}{3}+\frac{2(3a-1)}{3}$$ $$\mathrm{for}a⩾1\mathrm{we}\mathrm{get}$$ $$f\left(a\right)=\frac{2(3a-1)}{3}$$

Commented byliberty last updated on 19/Nov/20

$$byparts\int vdu=vu-\int udv?$$

Commented byMJS_new last updated on 19/Nov/20

$$\int u^{\prime }v=uv-\int {uv}^{\prime }$$

Commented byliberty last updated on 19/Nov/20

$$haha..{it}^{\prime }ssamesir$$

Commented byMJS_new last updated on 19/Nov/20

$$\mathrm{yes}$$

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