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

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?

#### Solutions Collecting From Web of "How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$?"

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. ðŸ˜‰

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.

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: