
#1
Dec2003, 11:46 AM

P: 5

Ssec^3(w)dw
it is very easy if you use integration by parts, but you need to pay atention! how about use a partial fraction ? if you like to have a hard work!! Someone agree with me? 



#2
Dec2003, 06:49 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

Actually, the way I would do that is to write sec^{3}w as
1/cos^{3}(w) and sec^{3}(w)dw as cos(w)dw/cos^{4}(w)= cos dw/(1 sin^{2}(w))^{2}, make the substitution u= sin(w) so the integral becomes ∫ du/(1u^{2})^{2} and THEN use partial fractions. Is that what you meant? 


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