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[lucid blaze]

Find the indefinite integral by transforming the variable

[upper bane]

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[brazen siren]

i suppose we can take out root 2 common from denominator thus forming a derivate of a known function

[zenith yoke]

bro

take the 9 outside the root

so it becomes 1/(3root(2/9 - x^2))

[zenith yoke]

no

[zenith yoke]

now just use

1/root(a^2-x^2) integral

is 1/a sin^-1(x/a)

[brazen siren]

yes so technically if we take out the root 2 in common

we get the same result

its 1/root(2).root(1-(3x/root2)^2)

[lucid blaze]

🙏