# How to evaluate $\int_0^a\frac{x^4}{(a^2+x^2)^4}dx$?

I have to evaluate

$$\int_0^a\frac{x^4}{(a^2+x^2)^4}\,{\rm d}x$$

I tried to substitute $x=a\tan\theta$ which then simplifies to $$\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta\, {\rm d}\theta$$.

Now,its quite hectic to solve this. Is there any other method out?

• I don’t know if there’s a quicker way, but you can write $cos^2x$ as $1-\sin^2x$ and then repeatedly use the $\sin$ reduction formulae – aidangallagher4 Sep 18 '18 at 11:38
• Maybe this helps: letting $x=at$ we get $$I= \frac{1}{a^3}\int_0^1 \frac{x^4}{(1+x^2)^4}dx= \int_0^1 \frac{1-\frac{1}{x^2}+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^4}dx = \int_0^1 \frac{\left(x+\frac{1}{x}\right)'}{\left(x+\frac{1}{x}\right)^4}dx +\int_0^1 \frac{dx}{x^2\left(x+\frac{1}{x}\right)^4}$$ I dont know any tricks for the last integral, but its pretty close to the original. – Zacky Sep 18 '18 at 11:39
• Why don't you choose the substitution $y = x^2+a^2$? – User123456789 Sep 18 '18 at 14:06
• There is a typo in your "${\rm}d\theta$" integral. It supposed to be $$\frac1{\color{red}{a^3}}\int_0^\frac\pi4\sin^4\theta\cos^2\theta\, {\rm}d\theta.$$ – Dinesh Shankar Sep 18 '18 at 17:23

It's not becoming "hectic" at all:

$$\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta d\theta=$$

$$\frac1{a^5}\int_0^\frac\pi4(\sin\theta\cos\theta)^2\sin^2\theta d\theta=$$

$$\frac1{8a^5}\int_0^\frac\pi4(\sin^22\theta)(1-\cos2\theta) d\theta=$$

$$\frac1{16a^5}\int_0^\frac\pi4(1-\cos4\theta)(1-\cos2\theta) d\theta=$$

$$\frac1{16a^5}\int_0^\frac\pi4(1-\cos4\theta-\cos2\theta+\cos4\theta\cos2\theta) d\theta=$$

$$\frac1{16a^5}\int_0^\frac\pi4(1-\cos4\theta-\cos2\theta+\frac12(\cos6\theta+\cos2\theta))d\theta=$$

$$\frac1{16a^5}\int_0^\frac\pi4(1-\cos4\theta-\frac12\cos2\theta+\frac12\cos6\theta)d\theta=$$

$$\frac1{16a^5}(\theta-\frac14\sin4\theta-\frac14\sin2\theta+\frac1{12}\sin6\theta)|_{0}^{\pi/4}=$$

$$\frac1{16a^5}(\frac\pi4-\frac13)=\frac1{192}(3\pi-4)$$

...and Mathematica gives the same result.

Once it is clear that the original integral is just $\frac{C}{a^5}$, with $$C=\int_{0}^{\pi/4}\sin^4(\theta)\cos^2(\theta)\,d\theta\stackrel{\theta\mapsto\varphi/2}{=}\frac{1}{2}\int_{0}^{\pi/2}\left(\frac{1-\cos\varphi}{2}\right)^2\left(\frac{1+\cos\varphi}{2}\right)\,d\varphi$$ one may simply expand everything and exploit $$\int_{0}^{\pi/2}\cos^{2m}(\varphi)\,d\varphi = \frac{\pi}{2\cdot 4^m}\binom{2m}{m},\qquad \int_{0}^{\pi/2}\cos^{2m+1}(\varphi)\,d\varphi = \frac{4^m}{(2m+1)\binom{2m}{m}}$$ which are crucial in many proofs of Wallis' product or Stirling's inequality, for instance.
Maybe not the most efficient approach, but for sure it is straightforward. It leads to $C=\frac{3\pi-4}{192}$.

HINT:

If you want to solve $\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta d\theta$ then

$1)$ Put $\theta=8t$ and then $\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta d\theta=\frac1{8a^5}\int_0^{2\pi}\sin^4{8t}\cos^2{8t} dt$

$2)$ Put $z=e^{it}$ and after some substitutions you will have an integral of a rational function $\frac{P(z)}{Q(z)}$

Now you can find the poles of the function in the region surrounded by the unit circle and apply the residue theorem.

With the above substitution: $$\sin{t}=\frac{z-\frac{1}{z}}{2i}$$ $$\cos{t}=\frac{z+\frac{1}{z}}{2}$$

You can also find similar relations for $$\sin{8t},\cos{8t}$$

The change of variable $x=a\tan\theta$ give us

$$I=\int_0^a\frac{x^4}{(a^2+x^2)^4}\, dx=\frac{1}{a^3}\int_0^{\pi/4}\sin^4(\theta)\cos^2(\theta) \,d\theta.$$

Now write $\cos^2(\theta)=1-\sin^2(\theta)$, such that

$$I=\frac{1}{a^3}\int_0^{\pi/4}\left[\sin^4(\theta)-\sin^6(\theta)\right] \,d\theta.$$

Use then the reduction formula:

$$\int \sin^m(x)\,dx=-\frac{\cos(x)\sin^{m-1}(x)}{m}+\frac{m-1}{m}\int\sin^{-2+m}(x)\,dx.$$

Using this, one gets

$$\int_0^a\frac{x^4}{(a^2+x^2)^4}\,dx =\frac{3\pi-4}{192a^3}$$

$$I=\int_0^a\frac{x^4}{(a^2+x^2)^4}dx$$ firstly: $$\frac{1}{(a^2+x^2)^4}=\frac{1}{\left[a^2(1+\left(\frac{x}{a}\right)^2\right]^4}=\frac{a^{-8}}{\left[1+\left(\frac{x}{a}\right)^2\right]^4}$$ now we can use the substitution: $$x=a\tan(u)\,,dx=a\sec^2(u)du$$ so our integral becomes: $$I=\frac{1}{a^8}\int_0^{\pi/4}\frac{a^4\tan^4(u)}{\sec^8(u)}.a\sec^2(u)du=\frac{1}{a^3}\int_0^{\pi/4}\frac{\tan^4(u)}{\sec^6(u)}du=\frac{1}{a^3}\int_0^{\pi/4}\sin^4(u)\cos^2(u)du$$ which we can now split into two parts: $$I_1=\int_0^{\pi/4}\sin^4(u)du$$ $$I_2=\int_0^{\pi/4}\sin^6(u)du$$ now we just use the $$\sin(x)$$ reduction formula: $$\int\sin^n(x)dx=-\frac{1}{n}\sin^{n-1}(x)\cos(x)+\frac{n-1}{n}\int\sin^{n-2}(x)dx$$ which we can modify for our limits: $$\int_0^{\pi/4}\sin^n(x)dx=-\frac{1}{n}\left(\frac{1}{\sqrt{2}}\right)^n+\frac{n-1}{n}\int_0^{\pi/4}\sin^{n-2}(x)dx$$ and now it is easy to evaulate: $$I_1=-\frac{1}{16}+\frac{3}{4}\int_0^{\pi/4}\sin^2(x)dx$$ $$I_1=-\frac{1}{16}+\frac{3}{4}\left(-\frac{1}{4}+\frac{1}{2}\int_0^{\pi/4}dx\right)=-\frac{1}{4}+\frac{3\pi}{32}$$ so we can now move on to the second integral: $$I_2=-\frac{1}{48}+\frac{5}{6}I_1=-\frac{11}{48}+\frac{5\pi}{64}$$ so: $$I_1-I_2=\left(-\frac{1}{4}+\frac{3\pi}{32}\right)-\left(-\frac{11}{48}+\frac{5\pi}{64}\right)=\frac{3\pi-4}{192}$$ and we know that: $$I=\frac{I_1-I_2}{a^3}$$ therefore: $$I=\frac{3\pi-4}{192a^3}$$

In terms of trying to use another method, I guess you could try to use a form of the beta function. We know that: $$B(\alpha,\beta)=\int_0^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)dx=\frac{\Gamma\left(\frac{\alpha}{2}\right)\Gamma\left(1+\frac{\beta}{2}\right)}{2\Gamma\left(1+\frac{\alpha+\beta}{2}\right)}$$ and using another substitution: $$u=\frac{x}{2}$$ we can get: $$B(\alpha,\beta)=\int_0^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)dx=2\int_0^{\pi/4}\sin^{\alpha-1}(2u)\cos^{\beta+1}(2u)du=2\int_0^{\pi/4}\left(2\sin(u)\cos(u)\right)^{\alpha-1}\left(\cos^2(u)-\sin^2(u)\right)^{\beta+1}du$$ although this seems to be more trouble than just evaluating the integral using the reduction formula.

This is how you derive the reduction formula: $$\int\sin^n(x)dx=-\sin^{n-1}(x)\cos(x)-(n-1)\int(-\cos(x))\sin^{n-2}(x)\cos(x)dx$$$$=-\sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)(1-\sin^2(x))dx$$$$=-\sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)dx-(n-1)\int\sin^n(x)dx$$ so: $$n\int\sin^n(x)dx=-\sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)dx$$ finally: $$\int\sin^n(x)dx=-\frac{1}{n}\sin^{n-1}(x)\cos(x)+\frac{n-1}{n}\int\sin^{n-2}(x)dx$$

Or you could attempt to use series: $$I=\int_0^a\frac{x^4}{(a^2+x^2)^4}dx=\int_0^a x^4(a^2+x^2)^{-4}dx=a^{-8}\int_0^a x^4\left(1+\left(\frac{x}{a}\right)^2\right)^{-4}dx$$$$=\frac{a^{-8}}{24}\int_0^a x^4\sum_{n=0}^\infty\left(\frac{x}{a}\right)^n (-1)^n \frac{(n+4)!}{n!}dx$$$$=\frac{1}{24}\int_0^a\sum_{n=0}^\infty (-1)^nx^{n+4}a^{-(n+8)}\frac{(n+4)!}{n!}dx$$$$=\frac{1}{24}\sum_{n=0}^\infty\left[x^{n+5}\right]_{x=0}^a a^{-(n+8)}\frac{(-1)^n(n+4)!}{(n+5)n!}$$$$=\frac{1}{24a^3}\sum_{n=0}^\infty\frac{(-1)^n(n+4)!}{(n+5)n!}$$ although I seem to have made an error in the series somewhere as the sum does not converge

EDIT: coming back to the beta function, there is another version called the incomplete beta function which states: $$B(z;a,b)=\int_0^z x^{a-1}(1-x)^{b-1}dx=z^a\sum_{n=0}^\infty\frac{(1-b)_n}{n!(a+n)}z^n$$ but again it is just easier to use a reduction formula or break it down for yourself using the identity: $$\cos(2x)=\cos^2(x)-\sin^2(x)=2\cos^2(x)-1=1-2\sin^2(x)$$ therefore: $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$

• If you can see any mistakes please edit to show me – Henry Lee Sep 21 '18 at 17:48

This is a rather standard rational function integral, and we only have to decompose it into partial fractions.

To have lighter calculations, we'll set $T=a^2+x^2$. Then $$x^4=T^2-2a^2x^2-a^4=T^2-2a^2T+a^4$$ which yields the decomposition \begin{align} \frac{x^4}{(a^2+x^2)^4}&=\frac{T^2-2a^2T+a^4}{T^4}=\frac1{T^2}-\frac{2a^2}{T^3}+\frac{a^4}{T^4}\\ &=\frac1{(a^2+x^2)^2}-\frac{2a^2}{(a^2+x^2)^3}+\frac{a^4}{(a^2+x^2)^4}. \end{align} Now the integrals $\;I_n=\int\frac{\mathrm d x}{(a^2+x^2)^n}$ can be computed recursively, from \begin{cases} I_1=\dfrac1a\arctan \dfrac xa, \\ I_{n+1}=\dfrac1{2na^2}\dfrac x{(a^2+x^2)^n}+\dfrac{2n-1}{2na^2} I_n. \end{cases} The recurrence relation can be obtained with an integration by parts of $I_n$.


$$$$\left.\int_{0}^{a}{x^{4} \over \pars{a^{2} + x^{2}}^4}\,\dd x \,\right\vert_{\ a\ \not=\ 0} = {1 \over a^{3}}\int_{0}^{1}{x^{4} \over \pars{1 + x^{2}}^4} \,\dd x \label{1}\tag{1}$$$$

\begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{x^{4} \over \pars{1 + x^{2}}^4}\,\dd x} = \left. -\,{1 \over 6}\,\totald[3]{}{b}\int_{0}^{1}{x^{4} \over b + x^{2}}\,\dd x\,\right\vert_{\ b\ =\ 1} \\[5mm] = &\ \left. -\,{1 \over 6}\,\totald[3]{}{b}\int_{0}^{1} \pars{-b + x^{2} + {b^{2} \over b + x^{2}}}\,\dd x\,\right\vert_{\ b\ =\ 1} \\[5mm] = &\ -\,{1 \over 6}\,\totald[3]{}{b}\bracks{b^{3/2}\int_{0}^{1} {1 \over \pars{x/\root{b}}^{2} + 1}\,{\dd x \over \root{b}}}_{\ b\ =\ 1} \\[5mm] = &\ -\,{1 \over 6}\,\totald[3]{}{b}\pars{b^{3/2}\int_{0}^{1/\root{b}} {\dd x \over x^{2} + 1}}_{\ b\ =\ 1} = \left. -\,{1 \over 6}\,\totald[3]{\bracks{b^{3/2}\arctan\pars{b^{-1/2}}}}{b}\right\vert_{\ b\ =\ 1} \\[5mm] = &\ \left.{\pars{b + 3}\pars{3b - 1} \over 8b\pars{b + 1}^{3}} - {3\,\mrm{arccot}\pars{\root{b}} \over 8b^{3/2}} \,\right\vert_{\ b\ =\ 1} = \bbx{3\pi - 4 \over 192} \approx 0.0283 \end{align} $$\bbx{\left.\int_{0}^{a}{x^{4} \over \pars{a^{2} + x^{2}}^4}\,\dd x \,\right\vert_{\ a\ \not=\ 0} = {3\pi - 4 \over 192a^{3}}}$$

Hint: Use these identities $$\sin^2t=\dfrac{1-\cos2t}{2}~~~~~,~~~~~~\cos^2t=\dfrac{1+\cos2t}{2}$$

• Would you demonstrate how this could help? – Hazem Orabi Sep 18 '18 at 11:49
• $$\int_0^\frac\pi4 (\dfrac{1-\cos2t}{2})^2(\dfrac{1+\cos2t}{2}) d\theta$$ – Nosrati Sep 18 '18 at 11:51