solve the indefinite integral $$\int\limits_{}^{}\frac{ \sin6t }{ \sin3t }$$

Question

hartnn · Answer

sin 6t = 2 sin 3t cos 3t

anonymous · Answer

right i got that

hartnn · Answer

still any problem solving the integral ?

anonymous · Answer

umm yea i guess when I get to the substitution part

hartnn · Answer

ok, so you are left with 2 cos 3t 
and you are trying to substitute 3t = u ?

anonymous · Answer

yes and then what should du be? du=2dt?

hartnn · Answer

3t = u 
3 dt = du
so, you will put 
dt as du/3

anonymous · Answer

oh because it's kind of like reverse chain rule and you would need to find the antiderivative of 3t? is that why it's 3dt?

hartnn · Answer

yes, you can say that. 
or just take the derivative, 
u= 3t
du/dt = 3 
du = 3dt

anonymous · Answer

ok that makes even more sense.  I just needed some sort of explanation.  IT wasn't making sense

hartnn · Answer

ok, good! now you can solve it entirely, right ? 
don't forget to re-substitute back u =3t
and add +c in the end

anonymous · Answer

so it would be (2/3)sin3t+c

hartnn · Answer

yes, it is :)