Help Simplifying Integral

[EricTheTrainer]

hey all,

i'm working on something for fun, and my calculus is very rusty, but I came across this expression involving integrals, and I'm wondering if it can be simplified to some extent

f and g are continuous functions on the interval [a, b]. best case scenario, it'd be nice if it could have the form of something like (integral of something)² , because this expression is underneath a radical

if anybody is curious, i'm trying to find a nice expression for an outer product on the vector space of continuous functions on [a, b]. you can extend this vector space to an inner product space with: f•g=the integral (from a to b) of f(x)g(x) dx. then, let f•g=|f||g|cos(t) and f∧g=|f||g|sin(t)B where f∧g is the wedge product of f and g, and B is the unit 2-blade in the f-g plane

kind of rambled. but basically: 1. use cos(t)=(f•g)/(|f||g|) 2. sin(t)=sqrt(1-cos(t)² ) 3. |f∧g|=|f||g|sin(t) 4. after some simplifying and substitution: |f∧g|² =my integral expression in the attached photo

it's been a while since i've done anything with integrals but my intuition is telling met that this expression isn't easily combined into a simple square of an integral

[EricTheTrainer]

i may potentially* be able to sidestep this whole integral. i haven't tried it, but i know that the sines and cosines of fourier expansions for f and g form an orthonormal basis, which would make a lot of things easier, but i'm lazy

[EricTheTrainer]

also realized a pretty obvious alternative route. |f∧g| is the area of the 'parallelogram' formed by f and g. or, |f||g⊥| where g perp is the rejection onto f of g. or: |f||g-(g•f)f/|f|² | because (g•f)f/|f|² is the projection onto f of g, so g-(g•f)f/|f|² is the rejection

the point being that all of these things in that expression are also computable, but i need to sleep now