Is $x=\tan\theta$ the only way for this integral?

Question

Evaluate $$\int\frac{dx}{(1+x^2)^4}$$ Now I did solve it, but I used the mentioned substitution and after a lot of converting into double angles, I did it. But, it doesn't look like a good approach. Is there a better way? Thanks.

You can also use hyperbolic functions: $1+\sinh^2 x=\cosh^2 x$, $d \sinh x=\cosh x dx$ — Yuriy S, May 08 '16 at 10:13

score 5 · Accepted Answer · answered May 08 '16 at 10:10

5

The standard way for $I_n=\displaystyle\int\frac{\mathrm d\mkern1mu x}{(1+x^2)^n} $ is to use integration by parts, setting $$u=\frac1{(1+x^2)^n}, \quad \mathrm d\mkern1mu v=\mathrm d\mkern1mu x,$$ to obtain a recurrence relation between $I_n$ and $I_{n+1}$. This relation is initialised by $\, I_1=\arctan x$.

answered May 08 '16 at 10:10

Bernard

175,478

@marmistrz there are two functions involved when integrating by parts. – Kaster May 08 '16 at 10:12
Aa, sure, misunderstood the notation. Deleted my comment – marmistrz May 08 '16 at 10:17

marmistrz · Answer 2 · 2016-05-08T10:10:29.253

1

Try this trick, which will lead you to a recurrence formula

$$\int\frac{dx}{(1+x^2)^n} = \int\frac{1 + x^2 - x^2}{(1+x^2)^n} dx$$

Let me know if it's not enough for a hint.

edited May 08 '16 at 10:10

answered May 08 '16 at 10:08

marmistrz

1,345

score 0 · Answer 3 · answered May 08 '16 at 15:08

0

Just a guess (I haven't tried it) but could you change it into partial fractions?

answered May 08 '16 at 15:08

Hassan

19

This is probably best as a comment and not an answer. – Karl May 08 '16 at 17:24

score 0 · Answer 4 · answered May 08 '16 at 18:58

If you let $x=\tan\theta$ then $$\begin{align}\int\frac{dx}{(x^2+1)^4}&=\int\cos^6\theta\,d\theta\\ &=\frac1{32}\int\left(\cos6\theta+6\cos4\theta+15\cos2\theta+10\right)d\theta\\ &=\frac1{32}\left(\frac16\sin6\theta+\frac32\sin4\theta+\frac{15}2\sin2\theta+10\theta\right)+C\\ &=\left(\frac16\left(32\sin^5\theta-32\sin^3\theta+6\sin\theta\right)\cos\theta\right.\\ &\left.+\frac32\left(4\sin\theta-8\sin^3\theta\right)\cos\theta+\frac{15}2\left(2\sin\theta\right)\cos\theta+10\theta\right)+C\\ &=\frac1{32}\left[\left(\frac{16}3\frac{x^5}{(x^2+1)^{5/2}}-\frac{52}3\frac{x^3}{(x^2+1)^{3/2}}+22\frac x{(x^2+1)^{1/2}}\right)\frac1{(x^2+1)^{1/2}}+10\tan^{-1}x\right]+C\\ &=\frac16\frac{x^5}{(x^2+1)^3}-\frac{13}{24}\frac{x^3}{(x^2+1)^2}+\frac{11}{16}\frac x{(x^2+1)}+\frac5{16}\tan^{-1}x+C\end{align}$$ In the above I expanded $\cos^6\theta=\left(\frac{e^{i\theta}+e^{-i\theta}}2\right)^6$ by the binomial theorem and then used ran the Chebyshev polynomials of the second kind out to $$\begin{array}{rl}\sin(2\theta)=&2\sin\theta\cos\theta\\ \sin(3\theta)=&3\sin\theta-4\sin^3\theta\\ \sin(4\theta)=&4\sin\theta\cos\theta-8\sin^3\theta\cos\theta\\ \sin(5\theta)=&16\sin^5\theta-20\sin^3\theta+5\sin\theta\\ \sin(6\theta)=&32\sin^5\theta\cos\theta-32\sin^3\theta\cos\theta+6\sin\theta\cos\theta\end{array}$$ Then since $x=\tan\theta$, $\cos\theta=\frac1{\sqrt{x^2+1}}$, and $\sin\theta=\frac x{\sqrt{x^2+1}}$. Since the result was fairly complicated, I verified it with numerical quadrature.

Is $x=\tan\theta$ the only way for this integral?

4 Answers4