Evaluating $ \int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x$.

Question

$$ \int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x. $$ What I have tried: I tried writing denominator as $ \sin^4x+\cos^4x = 1-2\sin^2x\cos^2x $ and $ 2\sin^2x\cos^2x = \frac{1}{2}\sin^2(2x) $ so the integral becomes, $$ \int\frac{\sin x+\cos x}{1-\frac{\sin^2(2x)}{2}}\,\text{d}x. $$ Anyone, how do I solve this further?

Answer 1

Let $$\displaystyle I = \int\frac{\sin x+\cos x}{\sin^4x +\cos^4 x}dx = \int\frac{\sin x+\cos x}{1-2\sin^2 x\cos^2 x}dx$$

So we get $$I = 2\int\frac{\sin x+\cos x}{2-(\sin 2x)^2}dx = 2\int\frac{\sin x+\cos x}{\left(\sqrt{2}-\sin 2x\right)\cdot \left(\sqrt{2}+\sin 2x\right)}dx$$

So we get $$I = \frac{1}{\sqrt{2}}\int \left[\frac{1}{\sqrt{2}+\sin 2x}+\frac{1}{\sqrt{2}-\sin 2x}\right]\cdot (\sin x+\cos x)dx$$

So we get $$I = \frac{1}{\sqrt{2}}\int \left[\frac{1}{1+\sqrt{2}-(\sin x-\cos x)^2}+\frac{1}{\sqrt{2}-1+(\sin x-\cos x)^2}\right]\cdot(\sin x+\cos x)dx$$

Now Put $(\sin x-\cos x) =t\;,$ Then $(\sin x+\cos x)dx = dt$

Answer 2

I suggest that you instead split the integral as $$ \int\frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,dx=\int\frac{\sin x}{(1-\cos^2x)^2+\cos^4x}\,dx+\int\frac{\cos x}{\sin^4x+(1-\sin^2x)^2}\,dx, $$ and then let $u=\cos x$ and $u=\sin x$ in the respective integral. You will get a (somewhat nasty) integral of a rational function, but it is standard.

Edit According to your comment (and I agree), you found out that one get integrals like (after expanding the square) $$ \int\frac{1}{2u^4-2u^2+1}\,du. $$ Now, you need to factor the denominator, and this is where things get somewhat nasty. I write down the result (in real terms), $$ 2u^4-2u^2+1=2\Bigl(u^2+\sqrt{1+\sqrt{2}}u+\sqrt{2}/2\Bigr)\Bigl(u^2-\sqrt{1+\sqrt{2}}u+\sqrt{2}/2\Bigr). $$ Thus, there are constants $A$, $B$, $C$ and $D$ such that $$ \frac{1}{2u^4-2u^2+1}=\frac{Au+B}{u^2+\sqrt{1+\sqrt{2}}u+\sqrt{2}/2}+\frac{Cu+D}{u^2-\sqrt{1+\sqrt{2}}u+\sqrt{2}/2}. $$ When that is done, you integrate "as usual". You will typically get logarithms and arctans, depending on what the constants $A$, $B$, $C$ and $D$ turn out to be.

The solution by juantheron is (as often) a short cut.