Question

When n is even, integrals of the form ∫tan^m(x)sec^n(x) dx can be evaluated by factoring out sec^2(x)=1+tan^2(x) and using the fact that Dx tanx=sec^2(x). When m is odd, integrals of this form can be evaluated by factoring out tanxsecx and using the fact that Dx secx=secxtanx. Use this method to evaluate the following integral: ∫tan^(-3/2)(x) sec^4(x) dx

Answer 1

Is this for calc 2 or something? Man you have some tricky problems! :O lol
\[\large \int\limits \frac{\sec^4x}{\tan^{3/2}x}dx\]

So `n is even`, (the power on the secant). Ok ok ok that's good.

Answer 2

yes calc 2 unfortunately

Answer 3

We'll turn 2 of the powers on secant into tangents.\[\large \int\limits\limits \frac{\sec^2x\sec^2x}{\tan^{3/2}x}dx \qquad = \qquad \large \int\limits\limits \frac{(1+\tan^2x)\sec^2x}{\tan^{3/2}x}dx\]

Answer 4

It might be easier to see what's going on if we move stuff around a little.\[\large \int\limits \frac{1+\tan^2x}{\tan^{3/2}x}(\sec^2x\;dx)\]See how we have a bunch of ugly tangent stuff... and thennnnn we also have the derivative of tangent sitting in those brackets!!! :O
Is the U sub jumping out at you yet?? :D

Answer 5

ok, let me try to see what i get

Answer 6

Oh man I hated calc 2...
These types of problems were actually a lot of fun in my opinions. I love all kinds of integrals. Area between curves, trig subs.. all that fun stuff :) Partial fractions, long division, impropers, fun fun fun.

Power series is where it got very difficult for me :C

Answer 7

lol ok im having trouble with this

Answer 8

and yes, not liking calc 2, but need it for my major. hopefully i can learn to love integrals like you though :)

Answer 9

\[\large u=\tan x\]\[\large du=\sec^2x\;dx\]
\[\large \int\limits\limits \frac{1+\tan^2x}{\tan^{3/2}x}(\sec^2x\;dx) \qquad \rightarrow \qquad \large \int\limits\limits \frac{1+u^2}{u^{3/2}}(du)\]
Were you able to get this far?

Answer 10

yes

Answer 11

From here we'll split up the fraction,\[\large \int\limits\frac{1+u^2}{u^{3/2}}du \qquad = \qquad \int\limits \frac{1}{u^{3/2}}+\frac{u^2}{u^{3/2}}du\]Divide the terms, apply negative exponents so it's easier to read, and integrate! :)

Answer 12

ok will give it a try now

Answer 13

so then we get 2(u^2-3)/3u^(1/2)

Answer 14

is that right so far?

Answer 15

Hmmm I'm not sureeeeee what's going on there, why don't you have two terms? Are you being all fancy and combining them? XD lol

Answer 16

\[\large \int\limits\limits u^{-3/2}+u^{1/2}\;du\]Then apply the power rule to each term.

Answer 17

i got it...so I put tanx into my u and i got [2(tan^2(x)-3)]/[3sqrt tanx]

Answer 18

lol sorry...just started playing around with it...but i definitely should separate them in future ...

Answer 19

thanks so much for your help zepdrix :)

Answer 20

yay team \c:/