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How would I go about solving the following integral:

$$T(x) = \int_\sqrt{x}^0 \sqrt{1+\sec^2 p}\ \mathrm dp $$

I went ahead and started by flipping the bounds, but am confused on next steps, this doesn't look like an identity to me but think that's probably the path to solving this integral.

Thank you for the help

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    $\begingroup$ Please type out your questions in LaTeX $\endgroup$ Nov 20 '19 at 17:35
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    $\begingroup$ Some things: $T$ doesn't depend on $p$ but on $x$. Also, there is a closed form of $$\int \sqrt{1+\sec^2(p)}\,\mathrm dp$$ but it is not pretty $\endgroup$ Nov 20 '19 at 17:44
  • $\begingroup$ will need to use an integral table $\endgroup$ Nov 20 '19 at 17:46
  • $\begingroup$ @RyRytheFlyGuy You can use the following series of substitutions: $u=\sec^2(x)$, $s=\operatorname{arcsec}(u)$, $p=\tan(s/2)$ and then use partial fractions to get a very ugly closed form $\endgroup$ Nov 20 '19 at 17:48
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Hint:

$$\sec^2x = 1 + \tan^2x$$

Therefore, $$\int \sqrt{1 + \sec^2 x}\,\mathrm dx = \int \sec^2x\cdot\frac{\sqrt{2 + \tan^2x}}{1 + \tan^2x}\,\mathrm dx$$

Next, use the substitution $u = \tan x$.

As @maximilian-janisch mentions in the comment, the solution isn't pretty, but this is something to get you started.

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Decompose the integrand into two and then integrate separately,

$$T(x)=\int_{\sqrt x}^0 \sqrt{1+\sec^2 p} dp$$ $$=\int_{\sqrt x}^0 \left(\frac{\cos p }{\sqrt{1+\cos^2 p}}+\frac{\sec^2 p }{\sqrt{1+\sec^2 p}}\right)dp$$

$$= \int_{\sqrt x}^0 \frac{d(\sin p)}{\sqrt{2-\sin^2p}} + \int_{\sqrt x}^0 \frac{d(\tan p)}{\sqrt{2+\tan^2p}}$$

$$=-\sin^{-1}\left( \frac1{\sqrt2}{\sin\sqrt x}\right) -\sinh^{-1}\left(\frac 1{\sqrt{2}}{\tan\sqrt x}\right)$$

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For the antiderivative, another solution could be $$I=\int \sqrt{1+\sec^2 (p)}\, dp=\int\sec ^2(p)\frac{\sqrt{\tan ^2(p)+2} }{\tan ^2(p)+1}\,dp$$ Now, let $q=\tan(p)$ to get $$I=\int\frac{\sqrt{q^2+2}}{q^2+1}\,dq$$ Now, let $q=\sqrt{2} \tan (r)$ to make $$I=2 \int \frac{ \sec ^3(r)}{2 \tan ^2(r)+1}\,dr$$ Now, let $r=\sin ^{-1}(s)$ to make $$I=-2 \int \frac{ds}{(s^2-1)(s^2+1)}=\int\left(\frac{1}{s^2+1}+\frac{1}{2 (s+1)}-\frac{1}{2 (s-1)} \right)\,ds$$ that is to say $$I=\tan ^{-1}(s)+\frac{1}{2} \log \left(\frac{1+s}{1-s}\right)$$ and, back to $p$, $$I=\tan ^{-1}\left(\frac{\tan (p)}{\sqrt{\tan ^2(p)+2}}\right)+\frac{1}{2} \log \left(\frac{\sqrt{\tan ^2(p)+2}+\tan (p)}{\sqrt{\tan ^2(p)+2}-\tan (p)}\right)+C$$

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