Question

integral of this equation:

Answer 1

\[\int\limits_{?}^{?} x \div (\cos^2 (3x^2))\]

Answer 2

\[=\int\limits{xsec(3x^2)dx}\]let u=3x^2
du=6x

Answer 3

I did that for the equation.

Answer 4

so the integration becomes\[\frac{1}{6} \int\limits{\sec^2(u)du}\]

Answer 5

in my first post its sec^(3x^2)

Answer 6

sec^2(3x^2)*

Answer 7

I think I went wrong somewhere, here's my work:
10. integral x/(cos^2(3x^2))

U = (3x^2)
dU = 6x
rewrite x/(cos^2(3x^2))---> (1/6)(sec^2(3x^2))6x dx
(1/6)(sec(U))^2du
2nd U sub Take w = (secu)
integral (1/6)w^2 = (1/6)(w^3/3) = w^3/18
input u.....(secu)^3/18
input (3x^2).....(sec^3(3x^2))/18

Answer 8

\[=\frac{1}{6}\tan(u)+C\]

Answer 9

integral of sec^2x=tanx

Answer 10

exactly :P

Answer 11

why do substitution twice?

Answer 12

I couldn't figure out how to find the integral of sec^2x.

Answer 13

http://math2.org/math/integrals/tableof.htm

Answer 14

Thanks lalaly!

Answer 15

ur welcome:D

Answer 16

nice :P

Answer 17

ur nice:D

Answer 18

ik that lalaly XD

Answer 19

hehe