Question

integrate 1/sec^2xtanx

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ \sec^2xtanx }dx\]

Answer 2

\[\int\limits_{?}^{?} \frac{ \cos x }{ \sec x \tan x }=\int\limits_{?}^{?} \frac{ 1 }{ \sec x \tan x }* \cos x\] you can use integration by parts now, and sec x tan x is differentiation of sec x

Answer 3

umm how did you get that first integral, sorry I'm confused

Answer 4

or would I just turn the sec^2x into (1-tan^2x)?

Answer 5

\[\int\limits_{}^{}cotxcos^2xdx\]

Answer 6

is that basically the same thing?

Answer 7

I hate calculus!!!

Answer 8

sorry just needed to vent

Answer 9

it just disappeared?

Answer 10

I'm starting to wonder if maybe I did the original problem wrong to even get to this scenario.... the original integral was \[\int\limits_{}^{}\frac{ 188 }{ x^2\sqrt{16x^2-81} }dx\]

Answer 11

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

Answer 12

opppss. i forgot + c :)

Answer 13

from there I got this: \[47\int\limits_{}^{}\frac{ 1 }{ x^2\sqrt{x^2-(\frac{ 9 }{ 4 })^2} }\]

Answer 14

so then I did \[x=\frac{ 9 }{ 4 }\sec\]

Answer 15

that should be sec theta

Answer 16

and from there I got \[47\int\limits_{}^{}\frac{ 1 }{ (\frac{ 9 }{ 4 })^2\sec^2\theta(\frac{ 9 }{ 4 })\tan \theta }d \theta \]

Answer 17

somebody stop me if I'm screwing up lol

Answer 18

i have done it above @stottrupbailey
happy valentine's day , hahaha

Answer 19

so then I get \[47 (\frac{ 4 }{ 9 })^3\int\limits_{}^{}\frac{ 1 }{ \sec^2\theta \tan \theta }d \theta\]

Answer 20

And thus the current dilemma, the integral of sec^2tan

Answer 21

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...

Answer 22

sorry, i didnt see the original problem.. i just see the first posting

Answer 23

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 :