Another indefinite integral question?

  • Thread starter uman
  • Start date
  • #1
352
0
How to evaluate [tex]\int{\frac{\arccos{x}}{x^2}\,dx}[/tex]?
 

Answers and Replies

  • #2
Try the substition
[tex]x=\cos u,\,d\,x=-\sin u\,d\,u[/tex]
and then integrate by parts.
 
  • #3
412
2
Try the substition
[tex]x=\cos u,\,d\,x=-\sin u\,d\,u[/tex]
and then integrate by parts.
Perfect..

[tex]
x = cos(u)~;~du = -sin(u)
[/tex]
also,
[tex]
x^2 = cos^2(u)
[/tex]

Therefore we have,

[tex]
\int\frac{cos^{-1}(x)}{x^2}dx = I = \int\frac{u sin(u)}{cos^{2}(u)}du
[/tex]

[tex]
I = -\int u tan(u)sec(u)du
[/tex]

[tex]
I = -\left[u sec(u) - \int \frac{du}{du} \int sec(u)tan(u)du du\right]
[/tex]

[tex]
I = -\left[u sec(u) - \int sec(u)du\right]
[/tex]

[tex]
I = log_e|sec(u) + tan(u)| - u sec(u) + C
[/tex]

Then, you can replace the value of u with [itex]cos(x)[/itex] to get the required integral.
 
  • #4
arildno
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Blunder there, you are to substitute [itex]u=\cos^{-1}(x)[/itex]
 
Last edited:
  • #5
412
2
Blunder there, you are to substitute [itex]u=\cos^{-1}(x)[/itex]
[itex]u = \cos^{-1}(x)~;~x = \cos(u)[/itex].. it's the same thing within a restricted domain.
 
  • #6
arildno
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You wrote substitute u with cos(x), which is wrong.
 
  • #7
412
2
ohh.. u mean in the last part.. yeah.. i get it now... ur right :D
 

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