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I have some answer: $$\int\frac{dz}{\sqrt{1+z^2}}=\ln ( z + \sqrt {1 + z^2})+C$$ But I can't get the same expression. So here what I've done and got stuck. How can I get the same answer?

\begin{align} & \int\frac{dz}{\sqrt{1+z^2}}=\left\{z=\tan t, \ dz=\frac{dt}{\cos^2 t} \right\} = \int\frac{dt}{\cos^2 t\sqrt{1+\tan^2 t}} \\[10pt] = {} & \left\{\sqrt{1+\tan^2 t}=\frac{1}{\cos t}\right\}=\int\frac{\cos t \ dt}{\cos^2 t} \\[10pt] = {} & \left\{\cos t \ dt= d\sin t, \cos^2 t=1-\sin^2 t\right\}=\int\frac{d\sin t}{1-\sin^2 t} \\[10pt] = {} &\{\sin t=u\}=\int\frac{d u}{1-u^2}=-\int\frac{du}{u^2-1}=-\frac{1}{2} \int\left(\frac{1}{u-1}-\frac{1}{1+u}\right)\,du \\[10pt] = {} &-\frac{1}{2}(\ln |u-1|-\ln|u+1|)+C \end{align}

From here I don't understand how to get the answer above. Need help

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  • $\begingroup$ Trying "unsubstituting" the values $u = \sin t$ and $z = \tan t$. $\endgroup$ Sep 26, 2017 at 0:15
  • $\begingroup$ why? It's an obvious question in the tittle now $\endgroup$ Sep 26, 2017 at 0:17
  • $\begingroup$ @ioleg19029700 Well, I hope the last few revisions were not in conflict with your question. Please review to ensure they aren't against your interests. $\endgroup$ Sep 26, 2017 at 0:48

3 Answers 3

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You're almost there. Note that with $u=\sin(t)$ and $z=\tan(t)$, we have

$$\begin{align} -\frac12\left(\log(|u-1|)-\log(|u+1|) \right)&=\frac12\log\left(\left|\frac{1+\sin(t)}{1-\sin(t)}\right|\right)\\\\ &=\frac12\log\left(\left|\frac{(1+\sin(t))^2}{1-\sin^2(t)}\right|\right)\\\\ &=\log\left(\left|\frac{1+\sin(t)}{\cos(t)}\right|\right)\\\\ &=\log(|\sec(t)+\tan(t)|)\\\\ &=\log(|z+\sqrt{1+z^2}|) \end{align}$$

as was to be shown!

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  • $\begingroup$ Didn't you miss minus after first "="? May you explain how you get these equalities? $\endgroup$ Sep 26, 2017 at 0:29
  • $\begingroup$ No. I absorbed the minus sign and flipped the argument of the logarithm. $\endgroup$
    – Mark Viola
    Sep 26, 2017 at 0:30
  • $\begingroup$ Thank you! Great fact $\endgroup$ Sep 26, 2017 at 0:36
  • $\begingroup$ You're welcome! My pleasure. $\endgroup$
    – Mark Viola
    Sep 26, 2017 at 0:37
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A better substitution is $z=\sinh t$, $\;\mathrm d\mkern 1mu z =\cosh t\,\mathrm d\mkern 1mu t$: $$\int\frac{\mathrm d\mkern 1mu z}{\sqrt{1+z^2}}=\int\frac{\cosh t\,\mathrm d\mkern 1mu t}{\sqrt{1+\sinh^2t}}=\int\frac{\cosh t\,\mathrm d\mkern 1mu t}{\cosh t}=t=\operatorname {argsinh}z=\ln\bigl(z+\sqrt{z^2+1}\bigr).$$

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I'm not sure about how obvious this seems, but you can just make the direct substitution $z=x+\sqrt{1+x^2}$ in$$\int\frac {\mathrm dx}{\sqrt{1+x^2}}$$ Because$$\frac {\mathrm dz}{\mathrm dx}=1+\frac {x}{\sqrt{1+x^2}}=\frac {z}{\sqrt{1+x^2}}$$ This can be noted because the derivative of $\sqrt{1+x^2}$ is $x/\sqrt{1+x^2}$, and if you add one, you get$$1+\frac x{\sqrt{1+x^2}}=\frac {x+\sqrt{1+x^2}}{\sqrt{1+x^2}}$$which is just the derivative of the original function. Therefore, by substitution, we have the integral as$$\begin{align*}\int\frac {\mathrm dx}{\sqrt{1+x^2}} & =\int\frac {\mathrm dz}{\sqrt{1+x^2}}\frac {\mathrm dx}{\mathrm dz}\\ & =\int\frac {\mathrm dz}{\sqrt{1+x^2}}\frac {\sqrt{1+x^2}}{z}\\ & =\log z+C\end{align*}$$

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