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stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ 1 }{ \sec^2xtanx }dx\]
 one year ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{?}^{?} \frac{ \cos x }{ \sec x \tan x }=\int\limits_{?}^{?} \frac{ 1 }{ \sec x \tan x }* \cos x\] you can use integration by parts now, and sec x tan x is differentiation of sec x
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
umm how did you get that first integral, sorry I'm confused
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
or would I just turn the sec^2x into (1tan^2x)?
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}cotxcos^2xdx\]
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
is that basically the same thing?
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
I hate calculus!!!
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
sorry just needed to vent
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
it just disappeared?
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
I'm starting to wonder if maybe I did the original problem wrong to even get to this scenario.... the original integral was \[\int\limits_{}^{}\frac{ 188 }{ x^2\sqrt{16x^281} }dx\]
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
int 1/sec^2xtanx dx = int cos^2 x cosx/sinx dx = int (1sin^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 usub 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
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
opppss. i forgot + c :)
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
from there I got this: \[47\int\limits_{}^{}\frac{ 1 }{ x^2\sqrt{x^2(\frac{ 9 }{ 4 })^2} }\]
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
so then I did \[x=\frac{ 9 }{ 4 }\sec\]
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
that should be sec theta
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
and from there I got \[47\int\limits_{}^{}\frac{ 1 }{ (\frac{ 9 }{ 4 })^2\sec^2\theta(\frac{ 9 }{ 4 })\tan \theta }d \theta \]
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
somebody stop me if I'm screwing up lol
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
i have done it above @stottrupbailey happy valentine's day , hahaha
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
so then I get \[47 (\frac{ 4 }{ 9 })^3\int\limits_{}^{}\frac{ 1 }{ \sec^2\theta \tan \theta }d \theta\]
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
And thus the current dilemma, the integral of sec^2tan
 one year ago

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
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...
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
sorry, i didnt see the original problem.. i just see the first posting
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
well, ur original problems is int ( 188/(x^2sqrt(16x^281)) 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^281), 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
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
from the equation : x = 9/4 secθ or secθ = 4x/9, reprsented in the figure : dw:1360828428409:dw
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
thus, the result is = 752/9 sinθ + c = 752/9 * sqrt(16x^2  81)/(4x) + c
 one year ago
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