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Question Number 112651 by bemath last updated on 09/Sep/20

$$\left(1\right)\underset{x\to \mathrm{a}}{\lim }(\mathrm{x}-\mathrm{a})\mathrm{cosec}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)?$$$$\left(2\right)\int \sqrt{{\mathrm{x}}^2+1}\mathrm{dx},\mathrm{by}\mathrm{using}{\mathrm{Euler}}^{\prime }\mathrm{s}$$$$\mathrm{substitution}$$

Answered by john santu last updated on 09/Sep/20

$$\left(★\right)by{Euler}^{\prime }ssubstitution$$$$let\sqrt{x^2+1}=x+q\Rightarrow x^2+1=x^2+2qx+q^2$$$$2qx+q^2=1\Rightarrow x=\frac{1-q^2}{2q}$$$$dx=\frac{-4q^2-2(1-q^2)}{4q^2}dq$$$$dx=\frac{-2q^2-2}{4q^2}dq=\frac{-q^2-1}{2q^2}dq$$$$soI=\int \sqrt{x^2+1}dx$$$$I=\int (\frac{1-q^2}{2q}+q).\left(\frac{-q^2-1}{2q^2}\right)dq$$$$I=-\int \left(\frac{q^2+1}{2q}\right)\left(\frac{q^2+1}{2q^2}\right)dq$$$$I=-\frac{1}{4}\int \left(\frac{q^4+2q^2+1}{q^3}\right)dq$$$$I=-\frac{1}{4}\int (q+\frac{2}{q}+q^{-3})dq$$$$I=-\frac{1}{4}(\frac{1}{2}{q}^2+2\ln \left(q\right)-\frac{1}{2q^2})+c$$$$I=-\frac{1}{8}{(\sqrt{x^2+1}-x)}^2-\frac{1}{2}\ln \mid \sqrt{x^2+1}-x\mid +\frac{1}{8{(\sqrt{x^2+1}-x)}^2}+c$$

Answered by bobhans last updated on 09/Sep/20

$$\left(⧫\right)\underset{x\to \mathrm{a}}{\lim }(\mathrm{x}-\mathrm{a})\mathrm{cosec}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)=\mathrm{L}$$$$\mathrm{L}=\underset{x\to \mathrm{a}}{\lim }\frac{(\mathrm{x}-\mathrm{a})}{\sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)}=\underset{x\to \mathrm{a}}{\lim }\frac{1}{\frac{\pi }{\mathrm{a}}\cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)}$$$$\mathrm{L}=\frac{\mathrm{a}}{\pi }\underset{x\to \mathrm{a}}{\lim }\frac{1}{\cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right)}=-\frac{\mathrm{a}}{\pi }$$

Answered by 1549442205PVT last updated on 10/Sep/20

$$2)\mathbf{Find}\int \sqrt{{\mathbf{x}}^2+1}\mathbf{dx}$$$$\mathrm{Put}\mathrm{x}=\mathrm{tant}\Rightarrow \mathrm{dx}=(1+{\tan }^2\mathrm{t})\mathrm{dt}.\mathrm{Then}$$$$\mathrm{F}=\int \sqrt{{\mathrm{x}}^2+1}\mathrm{dx}=\int \sqrt{1+{\tan }^2\mathrm{t}}(1+{\tan }^2\mathrm{t})\mathrm{dt}$$$$\mathrm{F}=\int \frac{\mathrm{dt}}{{\cos }^3\mathrm{t}}\mathrm{dt}=\frac{1}{2}\int \frac{1}{\mathrm{sint}}\times \frac{2\mathrm{sintcost}}{{\cos }^4\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{2}\int \frac{1}{\mathrm{sint}}\mathrm{d}\left(\frac{1}{{\cos }^2\mathrm{t}}\right)=\frac{1}{2}\times \frac{1}{{\mathrm{sintcos}}^2\mathrm{t}}$$$$-\frac{1}{2}\int \frac{1}{{\cos }^2\mathrm{t}}.{\left(\frac{1}{\mathrm{sint}}\right)}^{\prime }\mathrm{dt}=\frac{1}{2}\times \frac{1}{{\mathrm{sintcos}}^2\mathrm{t}}$$$$-\frac{1}{2}\int \frac{1}{{\cos }^2\mathrm{t}}.(-\frac{1}{{\sin }^2\mathrm{t}})\times \mathrm{costdt}$$$$=\frac{1}{2}\times \frac{1}{{\mathrm{sintcos}}^2\mathrm{t}}+\frac{1}{2}\int \frac{\mathrm{d}\left(\mathrm{sint}\right)}{{\cos }^2{\mathrm{tsin}}^2\mathrm{t}}$$$$\mathrm{Put}\mathrm{sint}=\mathrm{u}\mathrm{we}\mathrm{get}$$$$\frac{1}{2}\int \frac{\mathrm{d}\left(\mathrm{sint}\right)}{{\cos }^2{\mathrm{tsin}}^2\mathrm{t}}=\frac{1}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2(1-{\mathrm{u}}^2)}$$$$=\frac{1}{2}\int (\frac{1}{{\mathrm{u}}^2}+\frac{1}{1-{\mathrm{u}}^2})\mathrm{du}=-\frac{1}{2\mathrm{u}}-\frac{1}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2-1}$$$$=\frac{-1}{2\mathrm{u}}-\frac{1}{2}\times \frac{1}{2}\ln \mid \frac{\mathrm{u}-1}{\mathrm{u}+1}\mid .\mathrm{Therefore},$$$$\mathbf{F}=\frac{1}{2{\mathbf{sintcos}}^2\mathbf{t}}-\frac{1}{2\mathbf{sint}}-\frac{1}{4}\mathbf{ln}\mid \frac{\mathbf{sint}-1}{\mathbf{sint}+1}\mid +\mathbf{C}$$$$\mathbf{x}=\mathbf{tant}\iff {\mathbf{cos}}^2\mathrm{t}=\frac{1}{1+{\mathbf{x}}^2},\mathbf{sint}=\frac{\mid \mathbf{x}\mid }{\sqrt{1+{\mathbf{x}}^2}}$$$$\mathbf{F}=\frac{\mid \mathrm{x}\mid \sqrt{1+{\mathrm{x}}^2}}{2}-\frac{1}{4}\mathbf{ln}\mid \frac{\mid \mathbf{x}\mid -\sqrt{1+{\mathrm{x}}^2}}{\mid \mathbf{x}\mid +\sqrt{1+{\mathrm{x}}^2}}\mid +\mathbf{C}$$$$\mathrm{Also}\mathrm{see}\mathrm{question}112313$$

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