Interesting Integrals 8

Call the integration $\mathcal P$ and write it as:

$\mathcal P=-\int_0^{\pi/4}\dfrac{(\sin x+\cos x)\mathrm{d}x}{-4+\color{#D61F06}{(1-\sin 2x)}}=-\int_0^{\pi/4}\dfrac{(\sin x+\cos x)\mathrm{d}x}{-4+\color{#D61F06}{(\cos x-\sin x)^2}}$

Now substitute $\cos x-\sin x=u$ such that $-(\sin x+\cos x)\mathrm{d}x=\mathrm{d}u$ so that:

$\mathcal P=\int_1^0\dfrac{\mathrm du}{u^2-4}=\dfrac 14\int_1^0\left(\dfrac{1}{2-u}-\dfrac{1}{2+u}\right)\mathrm{d}u$

So that we finally get:

$\mathcal P=\dfrac 14\left[\ln\left|\dfrac{2-u}{2+u}\right|\right]_1^0=\dfrac{\ln 3}{4}$

Hence, $\Large 3-4=\boxed{-1}$

$\begin{aligned} I & = \int_0^\frac \pi 4 \frac {\sin x + \cos x}{3+\sin 2x} dx \\ & = \int_0^\frac \pi 4 \frac {\sqrt 2 \sin \left(x + \frac \pi 4 \right)}{3+\sin 2x} dx & \small \color{#3D99F6}{\text{By }\int_a^b f(x) \ dx =\int_a^b f(a+b-x) \ dx} \\ & = \int_0^\frac \pi 4 \frac {\sqrt 2 \cos x}{3+\cos 2x} dx & \small \color{#3D99F6}{\text{Let }u = \sin x \implies du = \cos x \ dx} \\ & = \int_0^\frac 1{\sqrt 2} \frac {\sqrt 2 \ du}{4-2u^2} & \small \color{#3D99F6}{\text{Let }\frac u{\sqrt 2} = \sin \theta \implies du = \sqrt 2 \cos \theta \ d\theta} \\ & = \frac 12 \int_0^\frac \pi 6 \frac {\cos \theta}{1-\sin^2 \theta} d \theta \\ & = \frac 12 \int_0^\frac \pi 6 \sec \theta \ d \theta \\ & = \frac 12 \ln \left( \frac {1+\sin \theta}{\cos \theta} \right) \bigg|_0^\frac \pi 6 \\ & = \frac 12 \ln \left( \frac {1+\frac 12}{\frac {\sqrt 3}2} \right) \\ & = \frac {\ln 3}4 \end{aligned}$

$\implies r - w = 3-4 = \boxed{-1}$