Question

The integral tan^3theta dtheta

Answer 1

\[\int\limits_{}^{}\tan ^3\theta d \theta\]

Answer 2

if we had a sec^2 it be nice :)

Answer 3

I wrote the equation so it's be better to see :)

Answer 4

Yea, I've been playing around with this one...but not really getting anywhere...

Answer 5

u = tan(t); tan-1(u) = t

du = sec^2(t) dt

dt = cos^2(t)

dt = cos^2(tan-1(u)) .... i wonder

Answer 6

Oh parts! I hadn't tried parts...hmmm

Answer 7

i think the trick is rather to split the tan^3 up into tan^2 and tan

Answer 8

tan^2 and tan lol ...cant type

Answer 9

Oh yea, I had split them first off...I figured it's just make it easier to work with.

Answer 10

tan^2 = (sec^2 - 1)

(sec^2 -1)tan dt maybe?

Answer 11

Tried that and did a u sub. u =\[\sec \theta\]

Answer 12

id have to write it on paper

Answer 13

You split them and then you have:
\[\int\limits \tan^2(x)\tan(x)dx=\int\limits (\sec^2(x)-1)\tan(x)dx=\int\limits \sec^2(x)\tan(x)dx-\int\limits \tan(x)dx\]
Let u=sec(x)
du=sec(x)tan(x)dx
\[\int\limits u du-\int\limits \tan(x)dx=\frac{u^2}{2}+\ln|\cos(x)|+C=\frac{\sec^2(x)}{2}+\ln|\cos(x)|+C\]

Answer 14

thats the route I was thinking of :)

Answer 15

Oh yea, okay that works!

Answer 16

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Answer 17

Thanks! Yes indeed, great job!

Answer 18

Thanks amistre :DDDD

Answer 19

\[\color{purple}{\text{WOW!}}\]

Answer 20

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Answer 21

Is the part that's cut off the integral of tanxdx?

Answer 22

Its not cut off on my screen but if you mean the end of the first line then yes.

Answer 23

okay cool