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how would I go about integrating this? It is a lecture exercise.
$$ \int x^2\sqrt{a^2+x^2}\,dx $$ I used a substitution of $x=a\tan\theta$ and ended up with $$\int a^4\tan^2\theta \sec^3\theta \,d\theta.$$ I was thinking by parts but then I would have to integrate a $\ln|\sec\theta+\tan\theta|$. Can't seem to think of the appropriate identity to collapse it. I am given the answer to verify but I don't know any mental algorithms to even begin doing inverse trigonometric substitution. The answer given is: $$ \frac{x}{8}(a^2+2x^2)\sqrt{a^2+x^2}-\frac{a^2}{8}\ln\left(x+\sqrt{a^2+x^2}\right) $$ Also, what are some things I should be thinking about first when looking at such questions when I need to substitute trigonometric functions into the x-variable? Should I attempt to make odd powers into even? How do I do so? Thank you very much for your help.

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Integrate by parts directly without substitutions

\begin{align} \int x^2\sqrt{a^2+x^2}dx = &\frac14 \int \frac x{\sqrt{a^2+x^2}}d[(a^2+x^2)^2]\\ = &\frac14 x{(a^2+x^2)^{3/2}}- \frac{a^2}4\int \frac{\sqrt{a^2+x^2}}{2x}d(x^2)\\ =& \frac14 x{(a^2+x^2)^{3/2}}- \frac{a^2}8x{\sqrt{a^2+x^2}}-\frac{a^4}8\int \frac{dx}{\sqrt{a^2+x^2}}\\ \end{align} where $\int \frac{dx}{\sqrt{a^2+x^2}}= \sinh^{-1}\frac xa$.

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Let us consider the integral $$\int\tan^2x\sec^3xdx=I.$$ Then using the identity $1+\tan^2x=\sec^2x,$ this becomes $$\int\sec^5xdx-\int\sec^3xdx.$$

Then integrating by parts, we find that $$\int\sec^5xdx=\sec^3x\tan x-3I.$$ Also we find that $$\int\sec^3xdx=\sec x\tan x-\int\sec^3xdx+\int\sec xdx.$$ Since $$\int\sec xdx=\log|\sec x+\tan x|+c,$$ we can easily find $$\int\sec^3xdx.$$

Finally, we get that $$4I=\sec^3x\tan x-\frac 12\left(\sec x\tan x+\log|\sec x+\tan x|\right)+C.$$

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  • $\begingroup$ Thank you very much for your help! I was doing something similar at the start, I got the sec5 - sec3, and even tried to get the same functional form on the right hand side, but I guess I'm just not good enough with trigonometric identities to do so, kept getting the wrong solution. If I might ask, if I replace the x back with the reverse substitution, I get the verified answer for the front part but not for the logged term. For the logged term, I get $log|\frac{\sqrt{a^2+x^2}}{a}+\frac{x}{a}|$. May I check if I am doing my reverse substitution correctly? $\endgroup$
    – Memiya
    Feb 16 at 17:21
  • $\begingroup$ @Memiya Your logarithmic term seems to be right. If you check the result of the other substitution I made using the hyperbolic functions, the log terms seem to agree. $\endgroup$
    – Allawonder
    Feb 17 at 8:54
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Another way to proceed with the original integral is to use the substitution $x=a\sinh y,$ so that the integral now is $$a^4\int \sinh^2y\cosh^2y dy=\frac{a^4}{4}\int\sinh^22y dy=\frac{a^4}{8}\int(1-\cosh 4y)dy=\frac{a^4}{8}\left(y-\frac14\sinh 4y\right)+k,$$ for some constant $k.$ Reverting the substitution now gives $$\frac{a^4}{8}\left[\log\left(\frac x a+\sqrt{\frac{x^2}{a^2}+1}\right)-\frac x a\left(1+\frac{x^2}{a^2}\right)-\frac{x^3}{a^3}\sqrt{1+\frac{x^2}{a^2}}\right]+C,$$ for some constant $C.$

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