Integrals of sine and cosine

$\begin{aligned} I & = \int_0^\frac \pi 4 \frac {\sin^3 x \cos^3 x}{\left(\sin^4 x + \cos^4 x\right)^2} dx & \small \color{#3D99F6} \text{Multiply up and down by }\sec^8 x \\ & = \int_0^\frac \pi 4 \frac {\tan^3 x \sec^2 x}{\left(\tan^4 x + 1\right)^2} dx & \small \color{#3D99F6} \text{Let }t = \tan x \implies dt = \sec^2 x\ dx \\ & = \int_0^1 \frac {t^3}{\left(t^4+ 1\right)^2} dt & \small \color{#3D99F6} \text{Let }u = t^4+1 \implies du = 4t^3 \ dt \\ & = \frac 14 \int_1^2 \frac 1{u^2} du \\ & = \frac 1{4u} \ \bigg|_2^1 \\ & = \frac 14 - \frac 18 \\ & = \frac 18 \end{aligned}$

Therefore, $a+b=1+8 = \boxed{9}$ .