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$$I=\int\frac{\sin x+\cos x}{\sin^2x+\cos^4x}\mathrm d x=?$$

I saw the related problem but that didn't help much. I attempted this question by splitting the numerator into $$I_1=\int\frac{\sin x}{\sin^2x+\cos^4x}\mathrm d x \text { and } I_2=\int\frac{\cos x}{\sin^2x+\cos^4x}\mathrm d x$$

For $I_1$, I set $t=\sin x$ and solved it the straightforward way. For $I_2$, I set $t=\cos x$ and again solved it the straightforward way.

My method got extremely lengthy in the end, while the question is supposed be decently okay. The answer that my book got is:

$$I=\frac 1{2\sqrt3}\ln\left|\frac{\sqrt3+t}{\sqrt3-t}\right|+\arctan t+C$$

for $t=\sin x-\cos x$. I have absolutely no clue how to get to such an easy result.

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  • $\begingroup$ It's cheating a little bit, but in light of the provided solution, I would try making the substitution $t = \sin x - \cos x$. $\endgroup$ – Theo Bendit Mar 9 '18 at 16:13
  • $\begingroup$ @TheoBendit Yep, I did try that myself but I never could successfully handle the power four cosine in the denominator. $\endgroup$ – Gaurang Tandon Mar 9 '18 at 16:14
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Here is a sketch only that should facilitate leading to the coveted result. To that end, we proceed.


First, noting that

$$\sin^2(x)+\cos^4(x)=\cos^2(x)+\sin^4(x)$$

we can write

$$\begin{align} \int\frac{\sin(x)+\cos(x)}{\sin^2(x)+\cos^4(x)}\,dx&=\int \frac{\cos(x)}{1-\sin^2(x)+\sin^4(x)}\,dx+\int \frac{\sin(x)}{1-\cos^2(x)+\cos^4(x)}\,dx\\\\ &=\left.\left(\int \frac1{v^4-v^2+1}\,dv\right)\right|_{u=\sin(x)}-\left.\left(\int \frac1{u^4-u^2+1}\,du\right)\right|_{u=\cos(x)} \end{align}$$


Next, we use partial fraction expansion to write

$$\begin{align} \frac{1}{x^4-x^2+1}&=\frac{1}{(x^2+\sqrt 3 x +1)(x^2-\sqrt 3 x+1)}\\\\ &=\frac{1}{2\sqrt3}\left(\frac{x+\sqrt3}{x^2+\sqrt 3x+1}-\frac{x-\sqrt3}{x^2-\sqrt 3x+1}\right)\\\\ &=\frac{1}{2\sqrt3}\left(\frac{x+\sqrt3/2+\sqrt3/2}{x^2+\sqrt 3x+1}-\frac{x-\sqrt3/2-\sqrt3/2}{x^2-\sqrt 3x+1}\right)\\\\ &=\frac{1}{4\sqrt3}\left(\frac{2x+\sqrt3}{x^2+\sqrt 3x+1}-\frac{2x-\sqrt3}{x^2-\sqrt 3x+1}\right)\\\\ &+\frac{1}{2}\left(\frac{1}{x^2+\sqrt 3x+1}-\frac{1}{x^2-\sqrt 3x+1}\right)\tag1 \end{align}$$


To finish, observe that the first two terms on the right-hand side of $(1)$ are perfect differentials and integration leads to logarithm terms. Integration of the last two terms lead to arctangent terms.

This way forward isn't pretty, but it is effective and tractable.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola May 27 at 18:36
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Since you have provided the hint in the answer $t = \sin x - \cos x$, we're going to use that. Thus, $\mathrm{d}t= (\cos x+\sin x)\mathrm{d}x$ which is already present in the numerator, so no issues there.

$t^2 = \sin^2x +\cos^2x -\sin2x$
$t^2 = 1 - \sin2x$
thus $\sin2x = 1- t^2$

In the denominator add and subtract $\cos^2x$ $\implies\sin^2x + \cos^2x -\cos^2x + \cos^4x\implies1 - \cos^2x\sin^2x$

The denominator will reduce to $1 - \sin^2x\cos^2x=1 - \sin^22x/4$

Break up the denominator as: $\frac{1}{(1-\sin2x/2)(1+\sin2x/2)}$

now put $\sin2x = 1 - t^2$

Now if you are comfortable with partial fractions the rest should be fairly easy! You can now integrate it with respect to $\mathrm{d}t$.

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If you want to do this the way that was suggested using $t=\sin x-\cos x$:

Then $$t^2=s^2-2sc+c^2\implies 4sc=2-2t^2$$ Also, $$t^4=1-4sc+4s^2c^2$$

So putting these results together gives $$4s^2c^2=t^4-2t^2+1$$

Meanwhile, the denominator of the integrand is $$s^2+c^4=s^2+c^2-s^2c^2=1-s^2c^2$$ $$=\frac 34+\frac 12t^2-\frac 14t^4$$

The numerator is just $dt$, so we end up with $$I=\int\frac{4dt}{3+2t^2-t^4}$$ $$=\int\frac{4dt}{(1+t^2)(3-t^2)}$$

This is very easily decomposed into $$\int\frac{1}{t^2+1}+\frac{1}{3-t^2}dt$$

The expected result follows immediately.

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  • $\begingroup$ I feel this was actually the expected method as well, as the result follows directly from here. Great and thanks! :) $\endgroup$ – Gaurang Tandon Mar 10 '18 at 2:18

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