Evaluate the definite integral $\int_0^a (a^2 + x^2)^\frac{5}{2} dx$

Question

I'm stuck on this definite integral problem. I need some constructive hint to proceed further.

$$\int_0^a (a^2 + x^2)^\frac{5}{2} dx$$

Substituting $$x = a \cot\theta,$$ I have converted this integral to $$\int_0^a (a \csc\theta)^5 dx.$$

Now I'm unable to proceed further. Kindly help

Answer 1

Hint:

Substitute $x=a\sinh(t)$.

$$a^6\int\cosh^6(t)\,dt=\frac{a^6}{32}\int(\cosh(6t)+6\cosh(4t)+15\cosh(2t)+40)\,dt.$$

Answer 2

Utilize the reduction formula $$\int_0^a \frac{x^n}{\sqrt{a^2+x^2}}dx=I_n=\frac{\sqrt2 a^6} n -\frac{(n-1)a^2}nI_{n-2} $$ to reduce the integral \begin{align} \int_0^a (a^2 + x^2)^\frac{5}{2} dx = &\int_0^a \frac{(a^2 + x^2)^3}{\sqrt{a^2+x^2}}dx\\ = &\int_0^a \frac{x^6+3a^2x^4+3a^4x^2+a^6}{\sqrt{a^2+x^2}}dx\\ =& \ \frac{67\sqrt2}{48}a^6+\frac5{16}a^6 I_0 \end{align} where $ I_0=\int_0^a\frac1{\sqrt{a^2+x^2}}dx=\sinh^{-1}1 $.

Answer 3

In $\theta$ the integral is $$a^6 \int_\frac\pi4^\frac\pi2 \csc^7 \theta \,d\theta.$$ Now apply the reduction formula $$\int \csc^m \theta \,d\theta = - \frac{1}{m - 1} \csc^{m - 2} \cot \theta + \frac{m - 2}{m - 1} \int \csc^{m - 2} \theta \,d\theta ,$$ which specializes for our limits to $$\int_\frac\pi4^\frac\pi2 \csc^m \theta \,d\theta = 2^{\frac{m}2 - 1} + \frac{m - 2}{m - 1} \int_\frac\pi4^\frac\pi2 \csc^{m - 2} \theta \,d\theta ,$$ three times to express the integral in terms of $$\int_\frac\pi4^\frac\pi2 \csc \theta \,d\theta = -\log |\csc \theta + \cot \theta| \Big\vert_\frac\pi4^\frac\pi2 = \operatorname{arsinh} 1 = \log(1 + \sqrt 2).$$

Performing these computations yields $$\boxed{a^6 \left(\frac{67 \sqrt 2}{48} + \frac{5}{16} \operatorname{arsinh} 1\right)}.$$