1
$\begingroup$

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

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?

$\endgroup$
  • 5
    $\begingroup$ Why not $t=\sin(2x)$? $\endgroup$ – clathratus Dec 18 '19 at 1:10
  • $\begingroup$ @clathratus the point of this exercise was to practice the t=tan(x/2) types of substitutions... but why t=sin(2x)? $\endgroup$ – Segmentation fault Dec 18 '19 at 1:13
  • 2
    $\begingroup$ Also, you made a typo : $$2\sin^2 x \cos^2 x = \tfrac12 \sin^2(2x).$$ $\endgroup$ – azif00 Dec 18 '19 at 1:14
  • 3
    $\begingroup$ Because if $t=\sin(2x)$ then $\frac12dt=\cos(2x)dx$ which you have on top $\endgroup$ – clathratus Dec 18 '19 at 1:14
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

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

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

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

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

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

| cite | improve this answer | |
$\endgroup$

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.