calc. trig substitution
posted by christine .
s integral
s 1/ [ (x^4) sq.rt(x^2+9)]
i know x=3tanx
sq.rt(x^2+9)= 3 secx
dx= 3/[cos^2(x)]
so far i know:
= 1/ (3tan^4(x)) 3secx cos^2(x)) dx
=1/ 81 [ (sin^4 (x)/cos^4 (x)) (1/cosx) (cos^2(x))]
then i'm not really sure what to do next
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