Here's the question you clicked on:
stottrupbailey
integrate 1/sec^2xtanx
\[\int\limits_{}^{}\frac{ 1 }{ \sec^2xtanx }dx\]
\[\int\limits_{?}^{?} \frac{ \cos x }{ \sec x \tan x }=\int\limits_{?}^{?} \frac{ 1 }{ \sec x \tan x }* \cos x\] you can use integration by parts now, and sec x tan x is differentiation of sec x
umm how did you get that first integral, sorry I'm confused
or would I just turn the sec^2x into (1-tan^2x)?
\[\int\limits_{}^{}cotxcos^2xdx\]
is that basically the same thing?
I hate calculus!!!
sorry just needed to vent
it just disappeared?
I'm starting to wonder if maybe I did the original problem wrong to even get to this scenario.... the original integral was \[\int\limits_{}^{}\frac{ 188 }{ x^2\sqrt{16x^2-81} }dx\]
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
from there I got this: \[47\int\limits_{}^{}\frac{ 1 }{ x^2\sqrt{x^2-(\frac{ 9 }{ 4 })^2} }\]
so then I did \[x=\frac{ 9 }{ 4 }\sec\]
that should be sec theta
and from there I got \[47\int\limits_{}^{}\frac{ 1 }{ (\frac{ 9 }{ 4 })^2\sec^2\theta(\frac{ 9 }{ 4 })\tan \theta }d \theta \]
somebody stop me if I'm screwing up lol
i have done it above @stottrupbailey happy valentine's day , hahaha
so then I get \[47 (\frac{ 4 }{ 9 })^3\int\limits_{}^{}\frac{ 1 }{ \sec^2\theta \tan \theta }d \theta\]
And thus the current dilemma, the integral of sec^2tan
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...
sorry, i didnt see the original problem.. i just see the first posting
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|
from the equation : x = 9/4 secθ or secθ = 4x/9, reprsented in the figure : |dw:1360828428409:dw|
thus, the result is = 752/9 sinθ + c = 752/9 * sqrt(16x^2 - 81)/(4x) + c