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

Question

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

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.

Answer 1

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 $\mathrm{d}t$.

Answer 2

Here is a sketch only that should facilitate leading to the coveted result. To that end, we proceed.

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First, noting that

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

we can write

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

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Next, we use partial fraction expansion to write

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

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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.

Answer 3

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.