Integration/calc work? what did i do wrong

  • #1
badtwistoffate
81
0
Int[1/(x^2+4)^2]
let x= 2 tan(q)
i get 1/(4 tan^2(q) + 4)^2
I get 1/ (16 tan ^4(q) + 32 (1-sec^2(q) + 16)
and now i think about doing a u-substituition with tan x=u, du =sec^2 (x) can i do that!?! :mad:
 

Answers and Replies

  • #2
badtwistoffate
81
0
do you just don't like my topics i want help in?
 
  • #3
whozum
2,221
1
[tex] \frac{1}{(4\tan(q)+4)^2} = \frac{1}{16\sec^4q} [/tex]
 

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