anonymous
  • anonymous
integrate 1/sec^2xtanx
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{}^{}\frac{ 1 }{ \sec^2xtanx }dx\]
anonymous
  • anonymous
\[\int\limits_{?}^{?} \frac{ \cos x }{ \sec x \tan x }=\int\limits_{?}^{?} \frac{ 1 }{ \sec x \tan x }* \cos x\] you can use integration by parts now, and sec x tan x is differentiation of sec x
anonymous
  • anonymous
umm how did you get that first integral, sorry I'm confused

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
or would I just turn the sec^2x into (1-tan^2x)?
anonymous
  • anonymous
\[\int\limits_{}^{}cotxcos^2xdx\]
anonymous
  • anonymous
is that basically the same thing?
anonymous
  • anonymous
I hate calculus!!!
anonymous
  • anonymous
sorry just needed to vent
anonymous
  • anonymous
it just disappeared?
anonymous
  • anonymous
I'm starting to wonder if maybe I did the original problem wrong to even get to this scenario.... the original integral was \[\int\limits_{}^{}\frac{ 188 }{ x^2\sqrt{16x^2-81} }dx\]
RadEn
  • RadEn
int 1/sec^2xtanx dx = int cos^2 x cosx/sinx dx = int (1-sin^2 x)cosx/sinx dx = int cosx/sinx dx - int sinxcosx dx = int cosx/sinx dx - int 1/2*sin2x dx case I : int cosx/sinx dx int by u-sub let u=sinx ---> du = cosx dx so, int cosx/sinx dx = int du/u = ln(u) = ln(sinx) case II : - int 1/2*sin2x dx = -(-1/4 cos2x) = 1/4 cos2x so, the total = ln(sinx) + 1/4 cos2x
RadEn
  • RadEn
opppss. i forgot + c :)
anonymous
  • anonymous
from there I got this: \[47\int\limits_{}^{}\frac{ 1 }{ x^2\sqrt{x^2-(\frac{ 9 }{ 4 })^2} }\]
anonymous
  • anonymous
so then I did \[x=\frac{ 9 }{ 4 }\sec\]
anonymous
  • anonymous
that should be sec theta
anonymous
  • anonymous
and from there I got \[47\int\limits_{}^{}\frac{ 1 }{ (\frac{ 9 }{ 4 })^2\sec^2\theta(\frac{ 9 }{ 4 })\tan \theta }d \theta \]
anonymous
  • anonymous
somebody stop me if I'm screwing up lol
RadEn
  • RadEn
i have done it above @stottrupbailey happy valentine's day , hahaha
anonymous
  • anonymous
so then I get \[47 (\frac{ 4 }{ 9 })^3\int\limits_{}^{}\frac{ 1 }{ \sec^2\theta \tan \theta }d \theta\]
anonymous
  • anonymous
And thus the current dilemma, the integral of sec^2tan
anonymous
  • anonymous
I'm confused by RadEn's explanation, can someone maybe explain what they did there? I got lost when he broke it up into subtracting two integrals...
RadEn
  • RadEn
sorry, i didnt see the original problem.. i just see the first posting
RadEn
  • RadEn
well, ur original problems is int ( 188/(x^2sqrt(16x^2-81)) dx in the frist step u have right let x = 9/4 secθ ---> x^2=9/16 sec^2 x dx = 9/4 secθtanθ dθ and for sqrt(16x^2-81), it can be sqrt(81sec^2 θ -81) = sqrt(81(sec^2 θ - 1) = 9sqrt(tan^2 θ) = 9tanθ so, ur integral can be : |dw:1360827737581:dw|
RadEn
  • RadEn
from the equation : x = 9/4 secθ or secθ = 4x/9, reprsented in the figure : |dw:1360828428409:dw|
RadEn
  • RadEn
thus, the result is = 752/9 sinθ + c = 752/9 * sqrt(16x^2 - 81)/(4x) + c

Looking for something else?

Not the answer you are looking for? Search for more explanations.