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Integtrating secant and tangent

  • Thread starter Aerosion
  • Start date
  • #1
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Homework Statement



(integrate)sec^3(2x)*tan(2x)dx

Homework Equations





The Attempt at a Solution



Okay, so first I tried to separate secant and tangent, to get (integrate)1/cos^3(2x) * sin(2x)/cos(2x)dx. The u would be cos(2x), the du would be -du=sin(2x)dx.

I subsitute these in the aforementioned equation, whcih leaves (integrate)1/u^3 x 1/u. The problem is, I don't think this is right.
 

Answers and Replies

  • #2
19
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u = 2x
du = 2 dx

[itex]\frac 12 \int \sec ^3 u \tan u du[/itex]

Now let [itex]v = \sec ^2 u[/itex] What is dv?
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
12,981
540

Homework Statement



(integrate)sec^3(2x)*tan(2x)dx

Homework Equations





The Attempt at a Solution



Okay, so first I tried to separate secant and tangent, to get (integrate)1/cos^3(2x) * sin(2x)/cos(2x)dx. The u would be cos(2x), the du would be -du=sin(2x)dx.

I subsitute these in the aforementioned equation, whcih leaves (integrate)1/u^3 x 1/u. The problem is, I don't think this is right.
There's the mistake.
 
  • #4
Gib Z
Homework Helper
3,346
5
Slight change to IMDereks post. I dont know wat dv is, cant be bothered working it out, i think my methods quicker. let v= sec u, split the integrand into sec^2 u (sec u tan u), that way the things in the brackets it dv/du, the integral is now integral: v^2 dv. simpler i think
 

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