At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

using the cos^2x + sin^2y identity

\[\int\limits_{1}^{4}\int\limits_{1}^{2}(x^2+y^2)/xy)dydx\]

why can't the x^2 + y^2 be set = 1

i think you are confusing a few things

was just kinda wondering when you can or can't use the x^2+y^2 = cos[x]^2 + sin[y]^2 = 1 identities

oh, i wrote it for the practice:)

:) heh

(..)^2 is not generally equativalent to the function trig^2(...)

You can use that identity to simplify integrals but you have to be carefull.

\[x^2+y^2\neq \cos^2x+\sin^2x\]

ah right but x^2/r^2 + y^2/r^2 = cos[x]^2 + sin[y]^2 = r^2 right?

no

Don't worry.

thta's some crazy cartesian-polar hybrid

hahaha

but its good to see you thinking outside the mathical box

ah i haven't read the chapter on polar stuff yet, so maybe i'll remedy things in the coming week :)

thank you all for the help!

welcome :)