Hints for integral

  • Thread starter sony
  • Start date
  • #1
104
0
Ok so after a substituion I have this: (I dont know how to proceed)

"integral of" (u-2) / sqrt(u^2 + 1) du

and by the way, can someone tell me what cot^3(tan^-1(x)) equals?

Thanks!
 

Answers and Replies

  • #2
tmc
289
1
sqrt(u^2+1) just screams for a trig substitution.
 
  • #3
104
0
Bah, I see it. I easily get confused when having to substitution i several steps :P

Thanks
 
  • #4
TD
Homework Helper
1,022
0
Well, it may be a bit easier if you split at first.

[tex]\int {\frac{{u - 2}}{{\sqrt {u^2 + 1} }}du} = \int {\frac{u}{{\sqrt {u^2 + 1} }}du} - \int {\frac{2}{{\sqrt {u^2 + 1} }}du} [/tex]

So the first one gives

[tex]\int {\frac{u}{{\sqrt {u^2 + 1} }}du} = \frac{1}{2}\int {\frac{{d\left( {u^2 + 1} \right)}}{{\sqrt {u^2 + 1} }}} [/tex]

Which is easy to integrate, without a trig subst.
Then of course, still the second part :smile:
 
  • #5
hotvette
Homework Helper
996
5
You might want to break it up into 2 integrals. The first one is very straightforward, the second one is easy with a trig substitution as tmc suggested.

EDIT: near simultaneous post w/ TD
 
Last edited:
  • #6
104
0
Ok, thanks. I have that one now.

Can you please hint me with one more:
I have the integral of (sqrt(1-x^2)) / x^4 dx

Which with trig.subs leads to:
integral of cot^2(y) *csc^2(y) dy

Again I'm stuck :P
 
  • #7
TD
Homework Helper
1,022
0
Correct, but remember that

[tex]\left( {\cot x} \right)^\prime = - \csc ^2 x[/tex]

:smile:
 

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