Question

Find the indefinite integral sec^2 and tan

Answer 1

\[\int\limits \sqrt{\tan(3x)} * (\sec(3x))^2 dx\]

Answer 2

I figured I'd do u = sec(3x)^2
du = tan(3x)

But that sqrt is in the way... I cannot have a sqrt(du) right?

Answer 3

try \(u=\tan(3x)\) to get what you want almost immediately

Answer 4

oh wait I did it backwards LOL.

Answer 5

you have it backwards
derivative of tangent is secant squared

Answer 6

There is no derivative of sec^2 :P too tired :P

Answer 7

so u = tan(3x)

du = sec(3x)^2 Does the chain rule apply here?

Answer 8

sec(3x)^2 dx

Answer 9

so then I have \[\sqrt{u}du --- u^{1/2} du\]

Answer 10

\[\frac{2(u)^{3/2}}{3} + c\]

Answer 11

hold on a sec

Answer 12

\[\frac{2(tan(3x))^{3/2}}{3} + c\]

Answer 13

\[u=\tan(3x),du=3\sec^2(3x)dx\] by the chain rule so \(\frac{1}{3}du=\sec^2(3x)dx\)

Answer 14

Ok so we do use the chain rule, ty.

Answer 15

so the end should be 2/6 then?

Answer 16

so you have
\[\frac{1}{3}\int\sqrt{u}du\] etc

Answer 17

anti derivative of \(\sqrt{u}\) is \(\frac{2}{3}u^{\frac{3}{2}}\) so you get
\[\frac{2}{9}u^{\frac{3}{2}}\]

Answer 18

yeah it is \(3\times 3=9\) and not 6 right? take a break it is saturday night
have a beer or whatever

Answer 19

I thought it was 1/3 * 1/2 ooopsies :P

Answer 20

Doing things too fast :P

Answer 21

\[\frac{2(\tan(3x))^{3/2}}{9} + c\]