# calc.- trig substitution

posted by
**christine** on
.

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