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I'm having trouble with following integral

$$ \int \frac{1}{\sqrt{1+4x^2+4y^2}} dy $$

that according to WolframAlpha is $$ \frac{1}{2} \ln( \sqrt{1+4x^2+4y^2} + 2y) + c $$

which I can verify but I can't figure out what substitution to apply or what functions to use for integration by parts in original integral. What would be your first step to attack this integral please? Thank you.

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  • $\begingroup$ Standard substitution for the form $a^2+ x^2$ is $x = a tan\theta$. That works here and gets you to $\int sec\theta\; d\theta$. $\endgroup$ – lulu Jul 23 '15 at 2:46
  • $\begingroup$ set $$\sqrt{4y^2+4x^2+1}=2y+t$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 23 '15 at 2:48
  • $\begingroup$ @Dr.SonnhardGraubner thank you, I've tried this but I can't get rid of y with this substitution. $\endgroup$ – sqxmn Jul 23 '15 at 3:18
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Notice, the standard formula $$\color{red}{\int\frac{dx}{\sqrt{a^2+x^2}}=\ln(x+\sqrt{a^2+x^2})+c}$$ For derivation, let $x=a\tan \theta \implies dx=a\sec^2\theta d\theta $ $$\int\frac{dx}{\sqrt{a^2+x^2}}=\int\frac{a\sec^2 \theta d\theta}{\sqrt{a^2+(a\tan\theta)^2}}=$$ $$=\int\frac{\sec^2 \theta d\theta}{\sec\theta}=\int\sec\theta d\theta=\ln(\sec\theta+\tan\theta)+c$$ $$=\ln(\sqrt{1+\tan^2\theta}+\tan\theta)+c$$ Now, substituting the value of $\tan \theta=\frac{x}{a}$ $$=\ln(x+\sqrt{a^2+x^2})+c$$

In the given integration, $x $ is treated as a constant. Now we have $$\int \frac{1}{\sqrt{1+4x^2+4y^2}} dy $$ $$ =\int \frac{1}{\sqrt{(1+4x^2)+(2y)^2}} dy $$ Let, $2y=t \implies 2dy=dt\ or\ dy=\frac{dt}{2}$ $$ =\frac{1}{2}\int \frac{dt}{\sqrt{(1+4x^2)+t^2}}+c$$ $$ =\frac{1}{2}\ln(t+\sqrt{(1+4x^2)+t^2})+c $$ $$ =\frac{1}{2}\ln(2y+\sqrt{(1+4x^2)+(2y)^2})+c$$ $$ =\frac{1}{2}\ln(\sqrt{1+4x^2+4y^2}+2y)+c $$

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  • $\begingroup$ Thank you! Very clear. I couldn't find this standard formula anywhere but I can verify it holds =). $\endgroup$ – sqxmn Jul 23 '15 at 3:05
  • $\begingroup$ Alright, Can I add its derivation at the top of answer? $\endgroup$ – Harish Chandra Rajpoot Jul 23 '15 at 3:06
  • $\begingroup$ I think that would be great. $\endgroup$ – sqxmn Jul 23 '15 at 3:09
  • $\begingroup$ You may see the derivation of the standard formula in the edited answer $\endgroup$ – Harish Chandra Rajpoot Jul 23 '15 at 3:19
  • $\begingroup$ Thank you! And thank you @lulu. Unfortunately I cannot upvote your answer yet. $\endgroup$ – sqxmn Jul 23 '15 at 3:24

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