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

itsmylife
 one year ago
Best 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

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

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

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

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

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

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

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

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

RadEn
 one year ago
Best ResponseYou've already chosen the best response.0int 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

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

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

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

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

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

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

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

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

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i didnt see the original problem.. i just see the first posting

RadEn
 one year ago
Best ResponseYou've already chosen the best response.0well, 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

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

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