Question

pleae check calculus answer @michele_laino @zepdrix

Answer 1

Oh, I didn't apply the chain rule. oops

Answer 2

hint:
we can write this:
\[\Large du = 3\sin \left( {3x} \right)dx\]

Answer 3

so, we get:
\[\Large \int {u\frac{{du}}{3}} = \frac{{{u^3}}}{6}\]

Answer 4

wait, wouldn't that be u^2, not u^3?

Answer 5

sorry, you are right! It is a typo:
\[\Large \int {u\frac{{du}}{3}} = \frac{{{u^2}}}{6}\]

Answer 6

okay, then just proceed by undoing the substitution, and solving for the integral

Answer 7

interval I mean

Answer 8

yes! That's right!