anonymous
  • anonymous
∫ sec^2 x tanx dx ------------------- (1+sec^2 x) ^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
are you sure it's over \[(1+\sec^2x)^2\]?
anonymous
  • anonymous
and not : \[(\sec^2x - 1)^2?\]
anonymous
  • anonymous
if not, then you can do it this way ^_^: 1) substitute sec^2x in the denominator with (1+tan^2x) and you'll get : \[\int\limits_{}^{}(\sec^2xtanx)/(1+1+\tan^2x)^2dx\] \[=\int\limits_{}^{}(\sec^2xtanx)/(2+\tan^2x)^2dx\] 2) let u = tanx and du = sex^2x dx: \[=\int\limits_{} (u)/(2+u^2)^2 du\] get (2+u^2)^2 up and you'll have : \[=\int\limits_{} u(2+u^2)^{-2}du\] now you can use integration by parts by letting the following as given : - f' = (2+u^2)^(-2) and g = u find f and g' then use the general equation which is: \[\int\limits_{}f'gdx = fg - \int\limits_{}fg'dx\] or partial fractions to solve this integration :) I hope I have given you hint, you can take it from here ^_^

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anonymous
  • anonymous
Correct me if I'm wrong :)

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