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I have this question given to be my professor, Solve $\int\frac{1}{\sqrt{x^2+1}}dx$

My way to solve it was quite straightforward.

$\int\frac{1}{\sqrt{x^2+1}}\,dx=\{ x = \sinh(t), \,dx = \cosh(t)\,dt \} = \int\frac{\cosh(t)}{\sqrt{\sinh(t)^2+1}}\,dt=\int 1\,dt = \sinh^{-1}(x)+C$

Which is quite straightforward and basically not difficult at all, the professor gave the same question and asked us to solve it using substitution $t=x+\sqrt{x^2+1}$

I am confused as I can not substitute that into the integral, what am I missing?

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    $\begingroup$ Hint: what is the derivative of $\,\ln(t)\,$? $\endgroup$ – Raymond Manzoni Mar 26 '17 at 17:47
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If $t= x+ \sqrt{1+x^2}$, then $$ \frac{dt}{dx} = 1 + \frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}} = \frac{t}{\sqrt{1+x^2}}, $$ so $$ \frac{dx}{\sqrt{1+x^2}} = \frac{dt}{t} .$$ Then the integral becomes $$ \int \frac{dx}{\sqrt{1+x^2}} = \int \frac{dt}{t} = \log{t}+C = \log{\left(x+\sqrt{1+x^2}\right)}+C $$

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  • $\begingroup$ Would combing the two results yields that $\log(x+\sqrt{1+x^2})=\arcsin(x)$ ? I don't have to prove anything about C's, do I? $\endgroup$ – Rab Mar 26 '17 at 18:05
  • $\begingroup$ @user2733996 what you have is only enough to conclude that they differ by a constant $\endgroup$ – bthmas Mar 26 '17 at 18:11
  • $\begingroup$ $\arg\sinh{x}=\log{(x+\sqrt{1+x^2})}$, since both sides have the same derivative and the same value at $x=0$, for example. $\endgroup$ – Chappers Mar 26 '17 at 18:12
  • $\begingroup$ If they are same, then where is the need to differentiate and do compensating integration ? $\endgroup$ – Narasimham Mar 26 '17 at 19:00
  • $\begingroup$ @Narasimham This is one way to prove they are the same. $\endgroup$ – Chappers Mar 26 '17 at 19:54
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$$x=tan(t)$$ $$dx=sec^2(t)dt$$

$$\int\frac{1}{\sqrt{tan(t)^2+1}}sec^2(t)dt$$

Recall Property: $$tan(t)^2+1 = sec^2(t)$$

So integral becomes:

$$\int\frac{sec^2(t)}{sec(t)}dt$$

Which is: $$\int\sec(t)dt$$ If you want to know how to solve this integral comment here:

This is a standard integral= $$ln|sec(t)+tan(t)|$$

Coming back to the substitution:

$$tan(t) = x$$ and $$sec(t) = \sqrt{x^2+1}$$

So integral is: $$ln|\sqrt{x^2+1}+x|$$

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    $\begingroup$ I think you forgot to read one part: "The professor gave the same question and asked us to solve it using substitution $t=x+\sqrt{x^2+1}$" $\endgroup$ – projectilemotion Mar 26 '17 at 18:19
  • $\begingroup$ Oh sorry ok. Sorry for misunderstanding $\endgroup$ – Sid Mar 26 '17 at 18:24
  • $\begingroup$ It's no problem. It is a valid answer, and it is worth mentioning. $\endgroup$ – projectilemotion Mar 26 '17 at 19:07
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Let $t= x+ \sqrt{1+x^2}$. Then $$\begin{align} \frac{dt}{dx}&= 1 + \frac{x}{\sqrt{1+x^2}} \\ & = \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}} \\ &= \frac{t}{\sqrt{1+x^2}}, \end{align}$$ so that $ \frac{dx}{\sqrt{1+x^2}} = \frac{dt}{t}$. Thus $$\begin{align} \int \frac{dx}{\sqrt{1+x^2}} &= \int \frac{dt}{t} \\ &= \log\lvert t\rvert+c \\ &= \log{\left\lvert x+\sqrt{1+x^2}\right\rvert}+c \\ &=\log{\left(x+\sqrt{1+x^2}\right)}+ c. \end{align}$$

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  • $\begingroup$ $\lvert x \rvert < \sqrt{1+x^2}$, so there's no need for taking absolute values. $\endgroup$ – Chappers Mar 26 '17 at 18:14
  • $\begingroup$ This is @Chappers answer nearly verbatim... $\endgroup$ – Benjamin Dickman Mar 26 '17 at 18:15
  • $\begingroup$ @BenjaminDickman True :/ $\endgroup$ – Shaun Mar 26 '17 at 18:19
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Edit: Below is an approach different from that of the professor's suggestion, and which does not use any hyperbolic trig functions for its substitutions. It is included simply as another way to broach the problem of finding the given antiderivative.


This is basically a rehashing of the earlier answer from Sid, but I always found it helpful, when dealing with trigonometric substitutions, to draw an actual right triangle:

enter image description here

You can see from this picture that $\sec(\theta) = \sqrt{x^2 + 1}$, that $\tan(\theta) = x$, and so $dx = \sec^2(\theta)d\theta$.

Your integral then becomes:

$$\int \frac{1}{\sqrt{x^2 + 1}}dx = \int \frac{\sec^2(\theta)}{\sec(\theta)}d\theta = \int \sec(\theta) d\theta = \log(\tan(\theta) + \sec(\theta)) + C$$

Converting back from $\theta$ to $x$ yields $\log(x + \sqrt{x^2 + 1}) + \text{Constant}$ as desired.

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    $\begingroup$ I think you forgot to read one part: "The professor gave the same question and asked us to solve it using substitution $t=x+\sqrt{x^2+1}$" $\endgroup$ – projectilemotion Mar 26 '17 at 18:19
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    $\begingroup$ @projectilemotion No, I read the OP and saw that there was already a full answer addressing this from Clappers. I just thought it worthwhile to include another method of solution, even if it is not precisely what was asked for. $\endgroup$ – Benjamin Dickman Mar 26 '17 at 18:28
  • $\begingroup$ Oh, I see. It is a good answer though (A unique approach). $\endgroup$ – projectilemotion Mar 26 '17 at 19:00
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    $\begingroup$ @Benjamin I use the same "triangle" approach to solving integrals with trignometric substitutions. Thanks for mentioning it. $\endgroup$ – Sid Mar 26 '17 at 20:24
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Both are same! There is this hyperbolic function identity between inverse hyperbolic function and logarithmic forms.

$$ \sinh^{-1}x == \log{\left(x+\sqrt{1+x^2}\right)} $$

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