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

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

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

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

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

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

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

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

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

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

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

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

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

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

RadEn
 2 years 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
 2 years 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
 2 years 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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