Question

evaluate the integral (tan3x)dx
i got: -1/3log(cos3x)

Answer 1

You got it right!!

(don't forget to include the constant of integration though)

Answer 2

integral tan3x dx

d(3x)=3dx

integral tan3x dx
=integral(1/3) tan3x d(3x)

tan = sin/cos
integral tanx dx = integral(1/cosx) ( -d(cosx) )
=-lncosx+C

integral tan3x dx
=integral(1/3) tan3x d(3x)
=(1/3)(-lncos3x)+C

you are right :D

Answer 3

it should be ln instead of log and you shoud have the absolute symbow to make sure than its not negatine sfter the ln and you should plus C at the end :D