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I'm trying to integrate by inverse trigonometric substitution, and I have an answer (though I'm pretty sure it is wrong), but it is not the answer that the solution provides. I need to check where I went wrong and what I should do differently. This is the question:$$ \int\frac{1}{(a^2-x^2)^2}dx$$ And this is what I got:$$ \frac{1}{a^4}\frac{1}{2}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)+C=\frac{x}{2a^4(a^2-x^2)}+\frac{\ln|\frac{a+x}{2a\sqrt{a^2-x^2}}|}{2a^3}+C$$ Whereas the solution should be:$$ \frac{x}{2a^2(a^2-x^2)}+\frac{1}{4a^3}\ln|\frac{x+a}{x-a}|+C$$ What I attempted was to do a by-parts once I got this: $$ \frac{1}{a^4}\int{\sec^3\theta d\theta} = \int\sec^2\theta\sec\theta d\theta $$ Thank you very much for your time and help.

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  • $\begingroup$ You need to edit a few things: (1) what was the inverse trig term you used for substitution? (2) Your last equation, has $\frac{1}{a^{4}}$ extra term on the left. (3) How did you get to that last line? It might be a good idea to show your solution since there might be something wrong there. $\endgroup$
    – Nima S
    Feb 17, 2021 at 3:08
  • $\begingroup$ Yes, I used $a=\sin{x}$. Sorry about the last line, it should have a $\frac{1}{a^4}$ on the right-hand side as well. I did by parts for that to reach the line that I got in the second equation from the top. Integration by parts yielded: $\frac{1}{a^4}[\sec\theta\tan\theta-\int\sec\theta d\theta+ln|\sec\theta\tan\theta|]$. Then I got the second equation from the top. For the exponent, should it not be 3 because when I replace $dx$ I need to write $a\cos\theta d\theta$, so it cancels out 1 exponent. $\endgroup$
    – Memiya
    Feb 17, 2021 at 4:45

1 Answer 1

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Substitution $x=a\sin\theta$ to proceed as follows \begin{align} \int\frac1{(a^2-x^2)^2}dx =&\frac1{a^3} \int \sec^3\theta d\theta = \frac1{2a^3} \int \frac{\sec\theta }{\tan\theta}d(\tan^2\theta)\\ =& \frac1{2a^3}\sec\theta\tan\theta + \frac1{2a^3} \int {\sec\theta }d\theta \\ = &\frac1{2a^3}\sec\theta\tan\theta + \frac1{2a^3} \ln|\sec\theta+\tan\theta|\\ = &\frac1{2a^3}\frac{\sin\theta }{\cos^2\theta}+ \frac1{2a^3} \ln|\frac{1+\sin\theta}{\cos\theta}|\\ =& \frac{x}{2a^2(a^2-x^2)} +\frac{1}{2a^3}\ln|\frac{x+a}{\sqrt{a^2-x^2}}|\\ =& \frac{x}{2a^2(a^2-x^2)} +\frac{1}{4a^3}\ln|\frac{(a+x)^2}{a^2-x^2}|\\ =& \frac{x}{2a^2(a^2-x^2)} +\frac{1}{4a^3}\ln|\frac{x+a}{x-a}| \end{align}

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  • $\begingroup$ Thank you very much for your help. Unfortunately, this is, for me, a lecture exercise for trigonometric substitutions so I must use $x=\sin\theta$. $\endgroup$
    – Memiya
    Feb 17, 2021 at 4:46
  • $\begingroup$ @Memiya - no problem. please see the edit for the substitution approach $\endgroup$
    – Quanto
    Feb 17, 2021 at 5:17
  • $\begingroup$ Thank you very much for your help, I realised that my mistake was in replacing $\theta$ with $x$ at the end. $\endgroup$
    – Memiya
    Feb 17, 2021 at 5:32

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