# Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$

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

I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).

I did:

$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\frac{\cos(2x)dx}{1-2\sin^2(2x)}$$

I tried applying the t=tan(y/2) (where y = 2x) but it became a mess so I figure this can be simplified further... help?

• Why not $t=\sin(2x)$? – clathratus Dec 18 '19 at 1:10
• @clathratus the point of this exercise was to practice the t=tan(x/2) types of substitutions... but why t=sin(2x)? – Segmentation fault Dec 18 '19 at 1:13
• Also, you made a typo : $$2\sin^2 x \cos^2 x = \tfrac12 \sin^2(2x).$$ – azif00 Dec 18 '19 at 1:14
• Because if $t=\sin(2x)$ then $\frac12dt=\cos(2x)dx$ which you have on top – clathratus Dec 18 '19 at 1:14

You may just apply $$t=\tan x$$, $$dx = \frac{dt}{1+t^2}$$ and the resulting expression is quite manageable,,

$$I=\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{1-\frac12\sin^2(2x)} = \int \frac {\frac{1-t^2}{1+t^2}\cdot\frac{dt}{1+t^2}}{1-\frac12\left(\frac{2t}{1+t^2}\right)^2}=\int \frac{1-t^2}{1+t^4}dt$$

Next, carry out the integral as follows,

$$I=-\int \frac{1 - \frac1{t^2}}{t^2+\frac1{t^2}}dt =-\int \frac{d(t + \frac1{t})}{(t+\frac1{t})^2-2} =\frac1 {\sqrt2} \coth^{-1} \left(\frac{t^2 + 1}{\sqrt2\>t}\right)+C$$

where $$(\coth^{-1}t)' = \frac1{1-t^2}$$ is used in the last step.

We know that \begin{align} \cos(2x) &= \cos^2(x) + \sin^2(x)\\ \sin(x) &= \dfrac{\tan(x)}{\sec(x)}\\ \sec^2(x) &= 1 + \tan^2(x) \end{align} So, $$\int\dfrac{\cos(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm dx \equiv \int\sec^2x\left(\dfrac{-(\tan(x) - 1)(\tan(x) + 1)}{\tan^4(x) + 1}\right)\,\mathrm dx$$ Let $$u = \tan(x)$$. So, $$\dfrac{\mathrm du}{\mathrm dx} = \sec^2(x)\to\mathrm dx = \dfrac{\mathrm du}{\sec^2(x)}$$. $$\implies\int\sec^2x\left(\dfrac{-(\tan(x) - 1)(\tan(x) + 1)}{\tan^4(x) + 1}\right)\,\mathrm dx\equiv-\int\dfrac{(u - 1)(u + 1)}{u^4 + 1}\mathrm du$$ Now, you should be able to factor the denominator and use partial fractions to get the final answer.

You are almost there. Set $$t=\sin(2x)$$ so that $$\frac12dt=\cos(2x)dx$$. Then, as @Azif00 noted, you integral would be $$\int\frac{1}{1-\frac12 t^2}\frac{dt}2=\int\frac{dt}{2-t^2}.$$ You can compute this with partial fractions or a substitution.

Alternatively, you could use $$u=\tan(x)$$ so that $$\sin(2x)=\frac{2u}{1+u^2},$$ $$\cos(2x)=\frac{1-u^2}{1+u^2},$$ and $$dx=\frac{2du}{1+u^2}.$$ Then we have your integral as $$2\int\frac{1}{1-\frac12(\tfrac{2u}{1+u^2})^2}\frac{1-u^2}{(1+u^2)^2}du\\ =2\int\frac{1-u^2}{u^4+1}du,$$ the rest of which I leave to you as an exercise.

Continues, we have $$\int{ \frac{1}{2 - t^2} dt }$$. assume $$t = \sqrt{2} \sin (y)$$, we differensial and get $$dt = \sqrt{2} \cos (y) dy$$ $$\int{ \frac{1}{2 - t^2} dt } = \frac{1}{2} \int{ \frac{\sqrt{2} \cos (y) dy}{1 - \sin ^2 (y) } dt } = \frac{\sqrt{2}}{2} \int{ \sec(y) dy } = \frac{\sqrt{2}}{2} \sec y \tan y = \frac{1}{2- t^2}$$ . back to question

$$\cos^4 x = \frac 18 (\cos 4x + 4\cos 2x + 3)\\ \sin^4 x = \frac 18 (\cos 4x - 4\cos 2x + 3)\\ \cos^4 x+ \sin^4 x = \frac 14(\cos 4x + 3)$$

$$\frac {4\cos 2x}{\cos 4x + 3}$$

That looks a little better.

Lets say $$\cos 4x = 1 - 2\sin^2 2x$$

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

Now we can do a u-subtitution

$$u = \sin 2x, du = 2\cos 2x$$

$$\int \frac {2}{4 - 2u^2}\ du$$

Separate into partial fractions

$$\frac 1{2\sqrt 2} \int \frac {1}{\sqrt 2-u} + \frac {1}{\sqrt 2+u} \ du$$

$$\frac 1{2\sqrt 2} (\ln (\sqrt 2+u)-\ln (\sqrt 2-u))$$

$$\frac 1{2\sqrt 2} (\ln (\sqrt 2+\sin 2x)-\ln (\sqrt 2-\sin 2x))$$