Stuck on definite integral problem due to inappropriate $\log$ I have this definite integral problem which I have solved correctly but I'm stuck in one of the steps. I have manipulated it but I think it's not feasible to solve it that way.
$$\int_0^a(a^2 + x^2)^\frac{5}{2} dx$$
I have substituted $$x = a\cot\theta$$
and then the integral will look something like this $$\int_0^a (cosec\theta)^{5} dx$$ 
I don't know how do I proceed further to solve this problem. Kindly help me solving this further.
 A: I would choose the substitution $x=a\cdot \tan \theta$, instead of $\cot\theta$. Also, it seems that you have not calculated $dx$ as a function of $\theta$ and $d\theta$ (that is, it is imperative to derivate the substitution you are using). Finally, $(a^2+a^2\tan^2\theta)^{5/2}=(a^2\sec^2\theta)^{5/2}=a^5 sec^5\theta$.
The integral $\int \sec^7\theta d\theta$ (which you will obtain after substituting correctly) can be solved as follows: 
First, you have to know how to integrate $\int \sec^3\theta d\theta$:
Using integration by parts, we can do $u=\sec\theta$ and $dv=\sec^2\theta d\theta$, obtaining $du=\sec\theta\tan\theta d\theta$ and $v=tan\theta$, and we have
$$\int sec^3\theta d\theta=\sec\theta\tan\theta-\int sec\theta(\tan^2\theta)d\theta$$
$$=\sec\theta\tan\theta-\int sec\theta(\sec^2\theta -1)d\theta$$
$$=\sec\theta\tan\theta-\int sec^3\theta d\theta+\int\sec\theta d\theta$$
Then,
$$2\cdot \int \sec^3 d\theta=\sec\theta\tan\theta+\int sec\theta d\theta=\sec\theta\tan\theta+\ln(\sec\theta + tan\theta)+C$$
and finally,
$$\int sec^3\theta d\theta=\dfrac{1}{2}\left(\sec\theta\tan\theta+\ln(\sec\theta + tan\theta)\right)+C$$
Knowing this, and using a similar method, you will be able to calculate $\int \sec^5 \theta d\theta$, and finally, $\int \sec^7\theta d\theta$.
A: When you want do integrate a function of $a^2+x^2$ you could let $x=$ any of $a\tan\theta, a\cot\theta, a\sinh\theta, \text{or} \,\,a\,\text{csch}\,\theta$. The substitution $x=a\sinh\theta$ has many advantages for this kind of problem. Then $a^2+x^2=a^2\cosh\theta$ and $dx=a\cosh\theta d\theta$ and so
$$\int_0^a\left(a^2+x^2\right)^{\frac52}dx=a^6\int_0^{\sinh^{-1}1}\cosh^6\theta d\theta$$
Since $\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right)$, $\sinh^{-1}1=\ln\left(1+\sqrt2\right)$ and since $\cosh(n+1)x=2\cosh nx\cosh x-\cosh(n-1)x$ we can run up Chebyshev polynomials pretty fast:
$$\begin{align}\cosh2x & =2\cosh^2x-1 \\ \cosh3x & =4\cosh^3x-3\cosh x \\ \cosh4x & =8\cosh^4x-8\cosh^2+1 \\ \cosh5x & =16\cosh^5x-20\cosh^3x+5\cosh x \\ \cosh6x & = 32\cosh^6x-48\cosh^4x+18\cosh^2-1\end{align}$$
We can solve this for $$\cosh^6x=\frac1{32}\left(\cosh6x+6\cosh4x+15\cosh2x+10\right)$$ Then
$$\begin{align}\int_0^a\left(a^2+x^2\right)^{\frac52}dx & =\frac{a^6}{32}\int_0^{\ln\left(1+\sqrt2\right)}\left(\cosh6\theta+6\cosh4\theta+15\cosh2\theta+10\right)d\theta \\ & =\frac{a^6}{32}\left.\left(\frac16\sinh6\theta+\frac32\sinh4\theta+\frac{15}2\sinh2\theta+10\right)\right|_0^{\ln\left(1+\sqrt2\right)}\end{align}$$
$\sinh0=0$ and $$\sinh6\ln\left(1+\sqrt2\right)=\frac12\left[\left(\sqrt2+1\right)^6-\left(\sqrt2-1\right)^6\right]=70\sqrt2$$
$$\sinh4\ln\left(1+\sqrt2\right)=\frac12\left[\left(\sqrt2+1\right)^4-\left(\sqrt2-1\right)^4\right]=12\sqrt2$$
$$\sinh2\ln\left(1+\sqrt2\right)=\frac12\left[\left(\sqrt2+1\right)^2-\left(\sqrt2-1\right)^2\right]=2\sqrt2$$ So
$$\int_0^a\left(a^2+x^2\right)^{\frac52}dx=\frac{a^6}{32}\left(\frac{70}6\sqrt2+18\sqrt2+15\sqrt2+10\ln\left(1+\sqrt2\right)\right)=\frac{a^6}{48}\left[67\sqrt2+15\ln\left(1+\sqrt2\right)\right]$$
