Question

What is the Indefinite integral of 1/(x^6sqrt(x^2-4)) using trigonometric substitution?

Answer 1

\[\int\limits \frac{ dx }{ x^6\sqrt{x^2-4} }\]

Answer 2

@anthonykanow put x=2sec theta

Answer 3

yes. and \[\sqrt{x^2-a^2}=2\tan(\theta)\]

Answer 4

\[\frac{ 2\sec(\theta)\tan(\theta) }{ 64\sec^6(\theta)2\tan(\theta) }\]= 1\[\frac{ 1 }{ 64 \sec^5(\theta) }\]

Answer 5

then, \[\int\limits \frac{ 1 }{\sec^5(\theta)}dtheta\] yes?

Answer 6

I mean \[1/64 \int\limits \frac{ 1 }{ \sec^5(\theta) } dtheta\]

Answer 7

Yeah then convert sec to cos in the numerator

Answer 8

@anthonykanow

Answer 9

can you proceed further? it is very easy now.

Answer 10

\[\int\limits_{}^{} \cos ^5 \theta d \theta \]

Answer 11

hello...

Answer 12

I know that 1/sec = cos

I am having trouble with the integral of cos^5(theta)

Answer 13

ok let me post a solution

Answer 14

Ok, thank you

Answer 15

sorry I think a better substitution is:
x=2 cosh(theta)

Answer 16

excellent, I didn't think of that. Thank you

Answer 17

@anthonykanow

Answer 18

@Michele_Laino please post your solution, its good to do a problem in various ways

Answer 19

ok! I write

Answer 20

Sorry I'm not able to use the editor. I try to log out and then to log in

Answer 21

do you have a math editor? then it will be very easy

Answer 22

here I am

Answer 23

\[x=2\cosh(\theta)\]
so we can write:
\[\frac{ 1 }{ 64 }\int\limits \frac{ d \theta }{ (senh(\theta))^{6} }\]

Answer 24

next?

Answer 25

I'm trying...

Answer 26

next is:
\[\frac{ 1 }{ 64 } \int\limits\limits \frac{ d \theta }{ (ch \theta) ^{6}}= \]
\[=\frac{ 1 }{ 64 }\int\limits d(th(\theta))(1-th(\theta))^{4}\]

Answer 27

then after developing the fourth power, we can integrate easily

Answer 28

@Princer_Jones