How to integrate $$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$

I tried the following approach: $$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\ = \frac{1}{2}\int \frac{1}{\sin^4x - \sin^2x + \frac{1}{2}} \,dx = \frac{1}{2}\int \frac{1}{(\sin^2x - \frac{1}{2})^2 + \frac{1}{4}} \,dx$$

The substitution $t = \tan\frac{x}{2}$ yields 4th degree polynomials and a $\sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach: $\sin^4x + \cos^4 x = (\sin^2 x + \cos^2x)(\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x = (1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)$
and then I tried substituting: $t = \sin x \cos x$ and got $$\int\frac{t\,dt}{2(1-2t^2)\sqrt{1-4t^2}}$$

Another way would maybe be to make two integrals: $$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{(1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)} \,dx = \\ \frac{1}{2}\int \frac{1}{1-\sqrt2\sin x \cos x} \,dx + \frac{1}{2}\int\frac{1}{1+\sqrt2\sin x \cos x} \,dx$$

... and again I tried $t = \tan\frac{x}{2}$ (4th degree polynomial) and $t=\sqrt2 \sin x \cos x$ and I get $\frac{\sqrt 2}{2} \int \frac{\,dt}{(1-t)\sqrt{1-2t^2}}$ for the first one.

Any hints?

  • $\begingroup$ Have you tried the Weierstrass substitution? $\endgroup$ – Lucian Jun 4 '14 at 19:44
  • $\begingroup$ Using the en.wikipedia.org/wiki/… will transform the integrand to the simpler $\frac1{a+b\cos4t}$. $\endgroup$ – Yves Daoust Jun 4 '14 at 19:48
  • 9
    $\begingroup$ Notice $$\sin^4 x + \cos^4x = 1 - 2\sin^2x\cos^2x = 1 - \frac12\sin^2(2x) = \frac14(3+\cos4x)$$ Introduce $t = \tan2x$, we get $$\int\frac{dx}{\sin^4 x + \cos^4x} = \int \frac{4}{3 + \frac{1-t^2}{1+t^2}}\frac{dt}{2(1+t^2)} = \int\frac{dt}{2+t^2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + \text{const.}$$ $\endgroup$ – achille hui Jun 4 '14 at 19:55

Simplify the denominator in the following way: $$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^2(2x)}{2}=\frac{1+\cos^2(2x)}{2}=\frac{2+\tan^2(2x)}{2\sec^2(2x)}$$ Hence, the integral you are dealing with is: $$\int \frac{2\sec^2(2x)}{2+\tan^2(2x)}\,dx$$ I guess the next step is pretty obvious now. ;)

  • $\begingroup$ May I ask how you got: $\frac{2+\tan^2(2x)}{2\sec^2(2x)}$ ? Did you use Wolfram Alpha, "try and error", a certain trick, a specific algorithm, ... ? I verified for myself that this is true but how to "come up" with something like this? $\endgroup$ – AltairAC Jun 13 '14 at 9:04
  • 1
    $\begingroup$ @Shirohige: I used only the basic trigonometric identities. First factor out $\cos^2x$ from the numerator and use that $\cos^2x=\frac{1}{\sec^2x}$. Then write $\sec^2x$ in the numerator as $1+\tan^2x$. $\endgroup$ – Pranav Arora Jun 13 '14 at 14:01

Another approach:

We have $$ \frac{1}{\sin^4x+\cos^4x},\tag1 $$ Multiply $(1)$ by $\dfrac{\tan^4x}{\tan^4x}$ we obtain $$ \frac{\tan^4x}{\sin^4x(1+\tan^4x)}=\frac{\sec^4x}{1+\tan^4x}=\frac{(1+\tan^2x)\sec^2x}{1+\tan^4x}.\tag2 $$ Letting $t=\tan x$, the integral turns out to be $$\eqalign { \int\frac{1+t^2}{1+t^4}\ dt&=\frac12\int\left[\frac{1}{t^2-\sqrt2t+1}+\frac{1}{t^2+\sqrt2t+1}\right]\ dt\\ &=\frac12\int\left[\frac{1}{\left(t-\dfrac1{\sqrt2}\right)^2+\dfrac34}+\frac{1}{\left(t+\dfrac1{\sqrt2}\right)^2+\dfrac34}\right]\ dt.\tag3 } $$ Using substitution $u=\dfrac{\sqrt3}2\left(t-\dfrac1{\sqrt2}\right)$ and $v=\dfrac{\sqrt3}2\left(t+\dfrac1{\sqrt2}\right)$, the integral in $(3)$ can easily be evaluated.

Addendum :

Another way to evaluate $\displaystyle\int\frac{1+t^2}{1+t^4}\ dt$ is dividing the integrand by $\dfrac{t^2}{t^2}$, we obtain $$\eqalign { \int\frac{1+\dfrac1{t^2}}{t^2+\dfrac1{t^2}}\ dt&=\int\frac{1+\dfrac1{t^2}}{\left(t-\dfrac1{t}\right)^2+2}\ dt. } $$ Now let $u=t-\dfrac1{t}\;\Rightarrow\;du=\left(1+\dfrac1{t^2}\right)\ dt$, the integral turns out to be $$ \int\frac{1}{u^2+2}\ du.\tag4 $$ The evaluation of the integral $(4)$ can follow @achillehui's comment.


Convert the exponential powers to multiple angles. From deMoivre's theorem, with $n\in\mathbb{N}$: $$ \begin{align} \left( e^{i \theta} \right)^{n} &= e^{i n\theta} \\ \left( \cos \theta + i \sin \theta \right)^{n} &= \cos n\theta + i \sin n\theta \end{align} $$ These intermediate formulas may help: $$ \begin{align} \cos 2\theta &= \cos^{2} \theta + \sin^{2} \theta \\ \sin 2\theta &= 2 \cos \theta \sin 2 \theta \\ \end{align} $$ Reduce the denominator $$ \sin ^4(x)+\cos ^4(x) = \frac{1}{4} (\cos (4 x)+3) $$ The primitive is $$ \int \frac{1}{\cos^{4}x + \sin^{4}x} \, dx = \int \frac{1}{\cos (4 x)+3} \, dx = \left(4 \sqrt{2}\right)^{-1}\arctan \left(\frac{\tan (2 x)}{\sqrt{2}}\right) $$

Here is a look at the integrand:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.