8

I am trying to integrate this function, which I got while solving $\int\frac{1}{\sin^4( x) + \cos^4 (x)}$:

$$\int \frac{1+x^2}{1+x^4}\mathrm dx$$

I think to factorise the denominator, and use partial fractions. But I cant seem to find roots of denominator. I also am unable to think substitution.

user1729
  • 31,015
jonsno
  • 7,521
  • See http://math.stackexchange.com/questions/426152/evaluating-int-0-infty-fracdx1x4 – lab bhattacharjee Jul 28 '16 at 15:57
  • 1
    @RafaelWagner It was already a decent question with decent formatting. How does putting the integral in block quotes and changing $dx$ to $\mathrm dx$ improve the question? The latter is purely a stylistic choice. And while using block quotes for emphasis is generally considered to be OK in certain situations, it's certainly not a good reason to bump an old question to the front page. – user232456 Mar 18 '17 at 18:14
  • @user232456 I like style. Better looking questions get more attention and editing is not an 'one person work'; better reputation people look at it and judge if it is a good edit or not. – R.W Mar 18 '17 at 18:22
  • @RafaelWagner The issue is that editing old questions pushes new questions off the front page. It can become extremely annoying when someone decides to frequently edit old questions just to make stylistic changes that they personally prefer. Furthermore, there is always the possibility that in the future block quotes will look completely different. So using them for what they're not intended for is not always the best idea. – user232456 Mar 18 '17 at 19:11

6 Answers6

13

One may write $$ \int \dfrac{1+x^2}{1+x^4}dx=\int \dfrac{1+\frac1{x^2}}{\frac1{x^2}+x^2}dx=\int \dfrac{\left(x-\frac1x\right)'}{\left(x-\frac1x\right)^2+2}\:dx $$ then one may conclude easily.

Olivier Oloa
  • 120,989
8

Hint: $$\int\frac{1+x^{2}}{1+x^{4}}dx=\frac{1}{2}\int\left(\frac{1}{x^{2}+\sqrt{2}x+1}+\frac{1}{x^{2}-\sqrt{2}x+1}\right)dx$$ $$=\frac{1}{2}\int\left(\frac{1}{\left(x+\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}}+\frac{1}{\left(x-\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}}\right)dx.$$

Marco Cantarini
  • 33,062
  • 2
  • 47
  • 93
4

Roots can be complex as well.

$x^4+1=0 \Leftrightarrow x \in \{e^{i k \pi / 4} \ ,\ k \in \{1,3,5,7 \} \}$.

You have 4 complex roots which are all different; hence, using denominator factorization and partial fractions as you mentioned should do the trick.

saleml
  • 563
  • Yes true, but then it would become cumbersome wouldn't it? Thanks. – jonsno Jul 28 '16 at 15:23
  • Nah the partial fractions will reduce it all down to sums of elementary functions. And more importantly it is a general method that would work for all functions of the form $\frac{P(x)}{Q(x)}$ for polynomials $P$ and $Q$. – mathreadler Jul 28 '16 at 16:45
4

$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$ $$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$ $$\sin^4 x + \cos ^4 x =\frac{1}{4}\left[(1- \cos2x)^2 +(1+ \cos2x)^2 \right] =\frac{1}{2}((1+ \cos^2 2x)$$ $$\implies 2 \int \frac{dx}{1+ \cos^2 2x}$$ $$= 2 \int \frac{\sec^2 2x}{\sec^2 2x+ 1}{dx}$$ $$t= \tan2x$$ $$\implies 2 \int \frac{dt}{2 + t^2}$$ $$\frac{1}{\sqrt2} \tan^{-1} \left(\frac{t}{\sqrt2}\right)$$

Aakash Kumar
  • 3,480
1

A more general approach would be to identify that we have a quotient of polynomials and realize we can do factorization and partial fraction decomposition. @Salem is on this track but does not mention how very general it is. It will work on any fraction of polynomials $$\frac{P(x)}{Q(x)} = \sum_{\forall i} \frac{P_i(x)}{Q_i(x)}$$ where for complex numbers and real polynomials these $P_i$ and $Q_i$ are ensured to be of maximum degree 2 and have real roots. So we are sure to get away with really nice terms to integrate once we have done the decomposition.

mathreadler
  • 25,824
1

Let $$I = \int\frac{1}{\sin^4 x+\cos^4 x}dx = \int\frac{1}{\sin^2 x\cos^2 x\left(\tan^2 x+\cot^2 x\right)}dx$$

$$I =\int\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x\left(\tan^2 x+\cot^2 x\right)}dx= \int\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+\left(\sqrt{2}\right)^2}dx$$

Now Put $\tan x-\cot x = t\;,$ Then $(\sec^2 x+\csc^2 x)dx=dt$

$$I = \int\frac{1}{t^2+(\sqrt{2})^2}dt = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right)+\mathcal{C}$$

So $$I = \int\frac{1}{\sin^4 x+\cos^4 x}dx = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)+\mathcal{C}.$$

Marco Cantarini
  • 33,062
  • 2
  • 47
  • 93
juantheron
  • 53,015