Question

solve for y' = tan^3(3x)sec(3x)

Answer 1

y' = dy/dx right?

so \[dy = \tan^3(3x)\sec(3x)dx\]

Answer 2

tan^3(3x)sec(3x)=tan^2(3x)tan(3x)sec(3x)=(1+sec^2(3x)))tan(3x)sec(3x)
put u=sec(3x)
du = 3 sec(3x) tan(3x)

Answer 3

\[
y=\int ( 1 +u^2) \frac {du}3
\]

Answer 4

hmm

Answer 5

ic hmm

Answer 6

\[1/3 [ u + u^3/3] +c\]

Answer 7

\[ u/3 + u^3/9 +c\]

Answer 8

sec(3x)/3 + sec^3(3x)/9 +c

Answer 9

I tihnk thats correct :D