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Integrate: $$\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx$$

Here's my attempt, using integration by parts in the first step:

\begin{align} \int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\int \:-\frac{x^2}{\left(9+x^2\right)^{1.5}}dx\right]^1_0 \\ &= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\left(\frac{x}{\sqrt{9+x^2}}-\ln \left(\frac{1}{3} \left|x+\sqrt{9+x^2}\right|\right)\right)\right]^1_0 \\ &= \left[\ln \left(\frac{1}{3}\left|x+\sqrt{x^2+9}\right|\right)\right]^1_0 \\ &= \ln \left(\frac{1+\sqrt{10}}{3}\right) \end{align}

I tried to solve this using substitution and I keep getting stuck at the same place. For the sake of understanding, I turned the definite integral into an indefinite one by remove the upper and lower bounds, since that's quite easy to do. I'm more concerned with the final result without the actual values. Here's my attempt:

\begin{align} \int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &\stackrel{{u = \frac{x}{3}}, dx = 3\,du} = \int\frac{3}{\sqrt{9u^2+9}}\,\mathrm{d}u \\ &= \int \dfrac{1}{\sqrt{u^2+1}}\,\mathrm{d}u \\ &\stackrel{{u=\tan(v)} \to \mathrm{d}u=\sec^2(v)\,\mathrm{d}v} = \int \frac{\sec^2(v)}{\sqrt{\tan^2(v)+1}}\,\mathrm{d}v \\ &= \int \sec(v)\,\mathrm{d}v \\ &= \int \frac{\sec(v) (\tan(v)+\sec(v))}{\tan(v)+\sec(v)}\,\mathrm{d}v \\ &= \int \frac{\sec(v) \tan(v)+\sec^2(v)}{\tan(v)+\sec(v)}\,\mathrm{d}v \\ &= \cdots \end{align}

Here is where I am stuck. I think that there should be another substitution here in order to get the final integral that can be solved, but I'm not sure which term to choose. I think it would have to be $w = w=\tan\left(v\right)+\sec\left(v\right)$, but I'm not $100 \%$ sure.

This really caught me off by surprise because it looks rather simple but boy is it tricky to do.

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  • $\begingroup$ You are almost there. With your $w$, what is $\frac {dw}{dv}$? Compare it with the numerator. $\endgroup$
    – player3236
    Commented Jan 18, 2021 at 14:17
  • $\begingroup$ Yup. Just take $w=\tan(v)+\sec(v)$, $dw=\sec^2(v)+\sec(v)\tan(v)$. Your integral reduces to $\int\frac{dw}w$, which should be rather simple to solve. $\endgroup$ Commented Jan 18, 2021 at 14:19
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    $\begingroup$ In MathJax and in LaTeX the code (\frac 1 2 + \frac 1 3) results in $\displaystyle (\frac 1 2 + \frac 1 3),$ whereas \left(\frac 1 2 + \frac 1 3 \right) results in $\displaystyle \left(\frac 1 2 + \frac 1 3\right),$ so including \left and \right is useful. But you repeatedly wrote \left(v\right) where either that or (v) would result in $(v).$ I think is a better not to add epiphenomenal complications to the code. It makes reading and editing the code easier. $\endgroup$ Commented Jan 18, 2021 at 18:24
  • $\begingroup$ . . . and you repeatedly wrote {\displaystyle\int} where just \int would work, and repeatedly write {=} rather than =, and those curly braces used in that way sometimes result in lack of proper spacing. $\qquad$ $\endgroup$ Commented Jan 18, 2021 at 18:32
  • $\begingroup$ @MichaelHardy Thanks for the suggestions! I'll definitely keep them in mind for any future questions! $\endgroup$
    – Ski Mask
    Commented Jan 18, 2021 at 18:32

4 Answers 4

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\begin{align} \int\frac{1}{\sqrt{x^2+9}}\,dx &=\int\sec(u)\,du\quad &(x=3\tan u)\\ &=\int \frac{\sec^2(u)+\tan(u)\sec(u)}{\tan(u)+\sec(u)}\,du\\ &=\int\frac{1}{s}\,ds&(s=\tan(u)+\sec(u))\\ &=\log(s)+C\\ &=\log(\tan(u)+\sec(u))+C\\ &=\log\bigg(\frac13(\sqrt{x^2+9}+x)\bigg)+C \end{align}

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Yes. Substitute $w = \tan v + \sec v $ then $dw = \sec v \tan v + \sec ^2 v dv$. So your integral transforms into $$ \int \frac{ dw}{w} = \log |w| = \log ( | \tan v + \sec v|)$$ and then it is easy to get back into desired expression.

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$$I=\int_0^1\frac{dx}{\sqrt{9+x^2}}=\frac13\int_0^1\frac{dx}{\sqrt{1+(x/3)^2}}$$ now we know that $\cosh^2\theta-\sinh^2\theta=1\Rightarrow 1+\sinh^2\theta=\cosh^2\theta$ so we can make the substitution $x/3=\sinh u\Rightarrow dx=3\cosh u\,du$ so lets do that: $$I=\frac13\int\limits_{\sinh^{-1}(0)}^{\sinh^{-1}(1/3)}\frac{3\cosh u}{\sqrt{1+\sinh^2u}}du=\int\limits_{\sinh^{-1}(0)}^{\sinh^{-1}(1/3)}du=\sinh^{-1}\left(\frac13\right)$$


using the fact that: $$\sinh^{-1}(x)=\ln\left(x+\sqrt{x^2+1}\right)$$ we can say that: $$I=\ln\left(\frac13+\sqrt{\frac19+1}\right)=\ln\left(\frac{1+\sqrt{10}}{3}\right)$$

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    $\begingroup$ That is quite a different approach and seems really interesting. Thanks for the answer! $\endgroup$
    – Ski Mask
    Commented Jan 18, 2021 at 18:31
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Firstly, a fairly standard integral that is indeed useful in your case:

$$\int\frac{1}{\sqrt{a^2+x^2}}dx=\text{arsinh}~\frac{x}{a}+C$$ Secondly, in your final integral we can factorise the integrand and reduce it to something very easy (if you know the integral of $\sec x$):

$${\displaystyle\int}\dfrac{\sec(v)\tan(v)+\sec^2(v)}{\tan(v)+\sec(v)}\,\mathrm{d}v={\displaystyle\int}\dfrac{\sec(v)(\tan(v)+\sec(v))}{(\tan(v)+\sec(v))}\,\mathrm{d}v=\int\sec(v)\mathrm{d}v$$ which is just $$\ln\lvert\sec(v)+\tan(v)\rvert +C$$ I hope that helps.

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