# Help with inverse trigonometric substitutions $\int x^2\sqrt{a^2+x^2}\,dx$.

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.

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 $$\int \frac{dx}{\sqrt{a^2+x^2}}= \sinh^{-1}\frac xa$$.

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

• 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? Feb 16 at 17:21
• @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. Feb 17 at 8:54

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 $$C.$$