# Indefinite integral $\int\frac{ \cos 2x}{\sin^4 x+\cos^4 x}\,dx$

Can someone help me with this integral:

$$\int\frac{\cos 2x}{\sin^4 x+\cos^4 x}\,dx$$

I don't understand why $\sin^4 x+\cos^4 x$ is not equal to $1$. Unfortunately, Wolfram doesn't show step by step solution. Thank you for your help.

• $\sin^2 +\cos^2 x=1$ but that says nothing about $\sin^3 +\cos^3 x$, $\sin^4 +\cos^4 x$, or higher powers. If you could factor $\sin^4 +\cos^4 x$ that would help. – Rory Daulton Dec 31 '15 at 12:27
• the Pythagorean Theorem states that $c^2=a^2+b^2$. Dividing both sides by $c^2$ we get $1=(\frac{a}{c})^2+(\frac{b}{c})^2$, or alternatively $1=\sin^2 x +\cos^2 x$. – John Joy Dec 31 '15 at 15:09

The reason why is that $$\left(\sin^2(x)+\cos^2(x)\right)^2=\sin^4(x)+\cos^4(x)+2\sin^2(x)\cos^2(x)=1^2=1$$ So, $$\sin^4(x)+\cos^4(x)=1-2\sin^2(x)\cos^2(x)=1-\frac 12 \sin^2(2x)$$ So $$I=\int \frac{\cos (2x)}{\sin^4 (x)+\cos^4 (x)}dx=\int \frac{\cos (2x)}{1-\frac 12 \sin^2(2x)}dx$$

Now, there is a beautiful thing here. I let you the pleasure of finding it and finishing the problem.

I am sure that you can take it from here.

• Thank you a lot. I've imagine that was that equality the problem. – G.S. Dec 31 '15 at 12:43
• You are welcome ! Happy New Year !!! – Claude Leibovici Dec 31 '15 at 13:19
• Claude, sorry for the Off topic, I just wanted to tell that I love your answer. I find them really kind and well explained! A happy new year to you too :D – Bman72 Dec 31 '15 at 13:26
• nice..................+1 – Bhaskara-III Dec 31 '15 at 13:44

Here are two substitutions that fit this integral well:

1) Since $$\frac{\cos 2x}{\cos^4x+\sin^4x}=\frac{1}{\cos^2x}\frac{1-\tan^2x}{1+\tan^4x}$$ we let $u=\tan x$. This will transform the integral into $$\int \frac{1-u^2}{1+u^4}\,du.$$ 2) Next, in the spirit of @juantheron, let $v=u+1/u$. This will transform the integral into $$\int \frac{1}{2-v^2}\,dv=\frac{1}{\sqrt{2}}\text{artanh}\frac{v}{\sqrt{2}}+C.$$ I leave it to you to do the substitutions back to the $x$ variable.

HINT:

$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\dfrac{(\sin2x)^2}2$$

Let $\sin2x=u$