King's Property Integration

We will first introduce the King's Property or the reflection property which is the following:

$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$

This is true because we can let $u=a+b-x$ , then

$\int_a^b f(a+b-x) dx = -\int_b^a f(u) du = \int_a^b f(u) du = \int_a^b f(x) dx$

So, in our problem we can see that

$I=\int_0^\frac \pi 2 \frac{\sin^3 x}{\sin x+\cos x} dx = \int_0^\frac \pi 2 \frac{\cos^3 x}{\sin x+\cos x} dx$

Adding these two together, we get

$2I=\int_0^\frac \pi 2 \frac{\sin^3 x+\cos^3 x}{\sin x+\cos x} dx$

$=\int_0^\frac \pi 2 \frac{(\sin x+\cos x)(\sin^2 x-\sin x\cos x+\sin^2 x)}{\sin x+\cos x} dx$

$=\int_0^\frac \pi 2 1-\sin x\cos x dx$

$=\int_0^\frac \pi 2 1 dx - \int_0^\frac \pi 2 \sin x \cos x dx$

$=\frac \pi 2 - \int_0^1 u du$ $(u =\sin x, du =\cos x dx)$

$=\frac \pi 2 - \frac 1 2$

$\Rightarrow I=\frac \pi 4 - \frac 1 4$

Similar solution as @ChengYiin Ong 's

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {\sin^3 x}{\sin x + \cos x} dx & \small \blue{\text{By reflection }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac {\sin^3 x}{\sin x + \cos x} + \frac {\cos^3 x}{\cos x + \sin x} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac {(\sin x+\cos x)\left(\sin^2 x - \sin x \cos x + \cos^2 x \right)}{\sin x + \cos x} dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(1 - \sin x \cos x \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(1 - \frac 12 \sin 2x \right) dx \\ & = \frac 12 \left[x + \frac 14 \cos 2x \right]_0^\frac \pi 2 \\ & = \boxed{\frac \pi 4 - \frac 14} \end{aligned}$