# Integrate $\int\frac{\sin x+\cos x }{(\sin x-\cos x)^{1/3}}\,\mathrm dx$.

Integrate $$\int\frac{\sin x +\cos x}{(\sin x-\cos x)^{1/3}}\,\mathrm dx.$$

What should I make $u$ and $\mathrm du$ equal? What should I do with this integration problem?

Integration by parts? I don't see how it's possible.

$u$-substitution? I don't know what to make $u$ or $\mathrm du$ equal to.

Let $u=\sin x-\cos x$, so $du=(\sin x+\cos x)\ dx$.
$$\int\frac{\sin x+\cos x}{(\sin x-\cos x)^{1/3}}dx=\int\frac1{u^{1/3}}du=\frac{3u^{2/3}}2+C$$
Now substitute $u$ back, you should get:
$$\frac{3u^{2/3}}2+C=\boxed{\frac32(\sin x-\cos x)^{2/3}+C}$$
Let $u=\sin x-\cos x$. Then $du=(\cos x+\sin x)\,dx$.