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Calculate $$\int\frac{1}{\sqrt{25+x^2}}dx$$ I tried using $u$ substitution and i got $$u=5x$$then$$u^2=25x^2$$and $$du=5dx$$ and my integral becomes $$\frac15\int\frac{dx}{\sqrt{u^2+2}}$$ but after this I am struck. Please help.

RAHUL
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    Why $2$? Also, you shouldn't have two $dx$ factors are the start. – J.G. Oct 16 '21 at 13:25
  • It should be the other way around: $x=5u$ and not $u=5x$. Then you end up with $$ \int {\frac{{du}}{{\sqrt {1 + u^2 } }}} . $$ – Gary Oct 17 '21 at 05:22

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If you do $\require{cancel}x=5\tan\theta$ and $\mathrm dx=5\sec^2\theta\,\mathrm d\theta$, then $\int\frac1{\sqrt{25+x^2}}\,\mathrm dx$ becomes$$\int\frac{\cancel5\sec^2\theta}{\sqrt{\cancel{25}(1+\tan^2\theta)}}\,\mathrm d\theta=\int\frac{\sec^2\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta=\int\sec\theta\,\mathrm d\theta.$$Now, use the fact that $\int\sec\theta\,\mathrm d\theta=\log|\sec(\theta)+\tan(\theta)|+C$.

José Carlos Santos
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Another approach to evaluate it would be to use the substitution $x=5\sinh u$ which gives us $\mathrm dx= 5\cosh u\,\mathrm du$.

This gives us $$\begin{align} \int \frac{\mathrm dx}{\sqrt{x^2+25}} &= \int\frac{5\cosh u}{\sqrt{25+25\sinh^2u}}\,\mathrm du \\ &= \int \frac{5\cosh u}{5\cosh u}\,\mathrm du\\ &= \int1\,\mathrm du \\ &= u+C \\ \int \frac{\mathrm dx}{\sqrt{x^2+25}} &= \operatorname{arcsinh} \Big(\frac x5\Big) +C \end{align}$$

Laxmi Narayan Bhandari
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${\sqrt{a^2+x^2}}\implies x=a \tan\theta \implies \mathrm dx=a\sec^2\theta\,\mathrm d\theta$

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

So

${\sqrt{5^2+x^2}} \implies x=5\tan\theta \implies \mathrm dx=5\sec^2\theta\,\mathrm d\theta$

$$\begin{align} \int \frac{\mathrm dx}{\sqrt{25+x^2}} &= \int\frac{5\sec^2\theta}{\sqrt{25+25\tan^2\theta}}\, \mathrm d\theta \\ &= \int \frac{5\sec^2\theta}{\sqrt{25(1+\tan^2\theta)}}\,\mathrm d\theta\\ &= \int \frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm d\theta\\&= \int \frac{\sec^2\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta\\&= \int \sec\theta\,\mathrm d\theta\\&= \ln |\sec \theta + \tan \theta| + C \end{align}$$

Ways to evaluate $\int \sec \theta \, \mathrm d \theta$

Now we have to convert back to $x$.

$$ \theta= \arctan\Big(\frac x5\Big)\\ \sec\theta= \frac{1}{\cos\theta}= \frac{\sqrt{25+x^2}}{5}\\ \tan\theta=\frac{x}{5}$$

$$\ln |\sec \theta + \tan \theta| + C_1 \\ = \ln \Big|\frac{\sqrt{25+x^2}}{5}\ + \frac{x}{5}\Big| + C_1 \\ = \ln(x+\sqrt{25+x^2})+C_2 $$ Here, $C_2 = C_1-\ln5$. We can remove the absolute sign by noticing that $\forall x\in \mathbb R $, $x+\sqrt{25+x^2}\gt 0$.

Thus, we conclude that $$\int\frac{\mathrm dx}{\sqrt{25+x^2}} = \ln(x+\sqrt{25+x^2})+C $$

tent123
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