The value of $$\int_0^{\pi/2} \frac{\sin^3 x}{\sin x + \cos x} dx$$ is:

We have to evaluate the definite integral

$$I=\displaystyle\int_{0}^{\pi/2}\frac{\sin^{3}x}{\sin x+\cos x}\,dx.$$

The standard trick for integrals of the form $$\int_{0}^{\pi/2}f(\sin x,\cos x)\,dx$$ is to use the property

$$\int_{0}^{\pi/2}f(x)\,dx=\int_{0}^{\pi/2}f\!\left(\frac{\pi}{2}-x\right)\,dx,$$

because replacing $$x$$ by $$\dfrac{\pi}{2}-x$$ merely reverses the limits.

Applying this to our integral, we write

$$I=\int_{0}^{\pi/2}\frac{\sin^{3}\!\bigl(\tfrac{\pi}{2}-x\bigr)}{\sin\!\bigl(\tfrac{\pi}{2}-x\bigr)+\cos\!\bigl(\tfrac{\pi}{2}-x\bigr)}\,dx.$$

Using the co-function identities $$\sin\!\left(\tfrac{\pi}{2}-x\right)=\cos x$$ and $$\cos\!\left(\tfrac{\pi}{2}-x\right)=\sin x,$$ the above becomes

$$I=\int_{0}^{\pi/2}\frac{\cos^{3}x}{\cos x+\sin x}\,dx.$$

Now add the original integral and this new expression:

$$2I=\int_{0}^{\pi/2}\left[\frac{\sin^{3}x}{\sin x+\cos x}+\frac{\cos^{3}x}{\sin x+\cos x}\right]dx =\int_{0}^{\pi/2}\frac{\sin^{3}x+\cos^{3}x}{\sin x+\cos x}\,dx.$$

We recall the algebraic identity for the sum of cubes:

$$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2}).$$

Putting $$a=\sin x$$ and $$b=\cos x,$$ we have

$$\sin^{3}x+\cos^{3}x=(\sin x+\cos x)\left(\sin^{2}x-\sin x\cos x+\cos^{2}x\right).$$

Substituting this into the integral gives

$$2I=\int_{0}^{\pi/2}\left(\sin^{2}x-\sin x\cos x+\cos^{2}x\right)\,dx.$$

Next we note the Pythagorean identity $$\sin^{2}x+\cos^{2}x=1,$$ so the integrand simplifies to

$$1-\sin x\cos x.$$

Therefore

$$2I=\int_{0}^{\pi/2}1\,dx-\int_{0}^{\pi/2}\sin x\cos x\,dx.$$

The first integral is straightforward:

$$\int_{0}^{\pi/2}1\,dx=x\Big|_{0}^{\pi/2}=\frac{\pi}{2}.$$

For the second integral we use the double-angle identity $$\sin x\cos x=\tfrac12\sin 2x.$$ Hence

$$\int_{0}^{\pi/2}\sin x\cos x\,dx =\frac12\int_{0}^{\pi/2}\sin 2x\,dx.$$

Let $$u=2x\;\Rightarrow\;du=2\,dx\;\Rightarrow\;dx=\frac{du}{2}.$$ When $$x=0,\;u=0;$$ when $$x=\tfrac{\pi}{2},\;u=\pi.$$ Thus

$$\frac12\int_{0}^{\pi/2}\sin 2x\,dx =\frac12\left(\frac12\int_{0}^{\pi}\sin u\,du\right) =\frac14\bigl[-\cos u\bigr]_{0}^{\pi} =\frac14\bigl[-\cos\pi+\cos 0\bigr] =\frac14\bigl[-(-1)+1\bigr] =\frac14(1+1)=\frac12.$$

Putting these results together, we find

$$2I=\frac{\pi}{2}-\frac12=\frac{\pi-1}{2}.$$

Finally, dividing both sides by 2, we obtain

$$I=\frac{\pi-1}{4}.$$

Hence, the correct answer is Option C.