How do I integrate $\int \frac{dx}{\sin^3 x + \cos^3 x}$? How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?
It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.
 A: \begin{align}
\sin^3(x)+\cos^3(x) & = (\sin(x)+\cos(x))(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)) \\
 &  \{\sin(x)+\cos(x) = \sqrt{2}\sin(x+\pi/4)\} \\ 
 & = \sqrt{2}\sin(x+\pi/4)(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)) \\
 & \{\sin(2x) = 2\sin(x)\cos(x)\} \\ 
 & =\sqrt{2}\sin(x+\pi /4)(\sin^2(x) + \cos^2(x) -\sin(2x)/2)) \\
 & =\sqrt{2}\sin(x+\pi /4)(1-\sin(2x)/2)) \\
 & \{u = x +\pi/4, \: \sin(2x)=-\cos(2u)\} \\
 & =(\sqrt{2}/2) \sin(u)(2+\cos(2u)) \\
 & \{\cos^2(u) = \frac{1}{2}(1 + \cos(2u)\} \\
 & =(\sqrt{2}/2) \sin(u)(1+2 \cos^2(u)) \\
\end{align}

\begin{align}
\ \int \frac{1}{\sin^3 (x) + \cos^3 (x)}dx & = \int{\frac{\sqrt{2}}{\sin(u)(1+2\cos^2(u))}du} \\
 & =\int{\frac{\sqrt{2}\sin(u)}{\sin^2(u)(1+2\cos^2(u))}du} \\
 & \{\cos(u)=t, \:-\sin(u) du = dt\} \\
 & =-\int{\frac{\sqrt{2}}{(1-t^2)(1+2t^2)}dt} \\ 
 & =-\sqrt{2} \int{\frac{1/6}{1-t}+\frac{1/6}{1+t}+\frac{2/3}{1+2t^2}dt} \\
\end{align}
A: Maybe this can help :
$$ \frac{2}{\sin x+\cos x}+\frac{\sin x+\cos x}{1-\cos x\sin x}=\frac{3}{\sin^3 x+\cos ^3 x}$$
and also
$$\sin x+\cos x=\sqrt 2 \sin \big( x+ \frac{\pi}{4} \big). $$
A: $\;\;\displaystyle\frac{1}{(\sin x+\cos x)(1-\sin x\cos x)}=A\left(\frac{\sin x+\cos x}{1-\sin x\cos x}\right)+B\left(\frac{1}{\sin x+\cos x}\right)$ 
where $1=A(\sin x+\cos x)^2+B(1-\sin x\cos x)=(A+B)+(2A-B)\sin x\cos x$.
Then $A=\frac{1}{3}$ and $B=\frac{2}{3}$,
so $\displaystyle\int\frac{1}{\sin^3 x+\cos^3 x}dx=\int\left(\frac{1}{3}\cdot\frac{\sin x+\cos x}{1-\sin x\cos x}+\frac{2}{3}\cdot\frac{1}{\sin x+\cos x}\right)dx$
$\displaystyle=\frac{1}{3}\int\frac{2\sin x+2\cos x}{2-2\sin x\cos x}dx+\frac{2}{3}\int\frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx$
$\displaystyle=\frac{1}{3}\int\frac{\sin x+\cos x}{1+(\sin x-\cos x)^2}dx+\frac{2}{3\sqrt{2}}\int\csc\left(x+\frac{\pi}{4}\right)dx$
$\displaystyle=\frac{1}{3}\arctan(\sin x-\cos x)+\frac{2}{3\sqrt{2}}\ln\big|\csc\left(x+\frac{\pi}{4}\right)-\cot\left(x+\frac{\pi}{4}\right)\big|+C$
$\displaystyle=\frac{1}{3}\arctan(\sin x-\cos x)+\frac{2}{3\sqrt{2}}\ln\left\vert\frac{\sqrt{2}-\cos x+\sin x}{\sin x+\cos x}\right\vert+C$
A: Maybe this will help:
$$=\int\frac{\tan^3x}{\tan^3x(\cos^3x+\sin^3x)}dx=\int\frac{\tan^3x}{\sin^3x(1+\tan^3x)}dx=\int\frac{\sec^3x}{1+\tan^3x}dx\\=\int\frac{(1+\tan^2 x)\sec x}{1+\tan^3x}dx$$
